20.暴力递归到动态规划（二）
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package com.learn.learn.tixi.twentieth;
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import java.util.HashMap;
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// https://leetcode.com/problems/stickers-to-spell-word/
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public class ArrToString {
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public static int minStickers1(String[] stickers,String target){
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int ans = process1(stickers,target);
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return ans==Integer.MAX_VALUE?-1:ans;
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}
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/*
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所有贴纸stickers，每一张贴纸都有无穷张
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target
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最少张数
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*/
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private static int process1(String[] stickers, String target) {
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if (target.length()==0){
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return 0;
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}
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int min = Integer.MAX_VALUE;
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// 每张贴纸被剪无数次还剩多少
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for (String first:stickers){
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String rest = minus(target,first);
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if (rest.length()!=target.length()){
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min=Math.min(min,process1(stickers,rest));
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}
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}
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return min+(min == Integer.MAX_VALUE?0:1);
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}
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private static String minus(String s1, String s2) {
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char[] str1 = s1.toCharArray();
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char[] str2 = s2.toCharArray();
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int[] count = new int[26];
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for (char cha:str1){
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count[cha-'a']++;
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}
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for (char cha:str2){
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count[cha-'a']--;
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}
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StringBuilder builder = new StringBuilder();
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for (int i = 0;i<26;i++){
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if (count[i]>0){
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for (int j=0;j<count[i];j++){
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builder.append((char) (i+'a'));
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}
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}
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}
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return builder.toString();
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}
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public static int minStickers2(String[] stickers,String target){
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int n = stickers.length;
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// 关键优化（用词频表替代贴纸数组）
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int[][]counts = new int[n][26];
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for (int i = 0;i<n;i++){
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char[]str=stickers[i].toCharArray();
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for (char cha:str){
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counts[i][cha-'a']++;
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}
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}
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int ans = process2(counts,target);
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return ans == Integer.MAX_VALUE?-1:ans;
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}
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// stickers[i] 数组，当初i号贴纸的字符统计 int[][] stickers ->所有的贴纸
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// 每一种贴纸都有无穷张
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// 返回搞定target的最少张数
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public static int process2(int[][] stickers,String t){
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if (t.length()==0){
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return 0;
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}
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char[] target=t.toCharArray();
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int[] tcounts= new int[26];
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for (char cha:target){
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tcounts[cha-'a']++;
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}
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int n = stickers.length;
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int min = Integer.MAX_VALUE;
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for (int i = 0;i<n;i++){
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int[] sticker = stickers[i];
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// 最关键优化（重要剪枝，这一步也很贪心）
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if (sticker[target[0]-'a']>0){
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StringBuilder builder = new StringBuilder();
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for (int j=0;j<26;j++){
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if (tcounts[j]>0){
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int nums = tcounts[j]-sticker[j];
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for (int k = 0;k<nums;k++){
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builder.append((char)(j+'a'));
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}
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}
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}
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String rest = builder.toString();
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min = Math.min(min,process2(stickers,rest));
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}
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}
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return min + (min == Integer.MAX_VALUE?0:1);
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}
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public static int minStickers3(String [] stickers,String target){
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int n=stickers.length;
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int[][] counts = new int[n][26];
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for (int i = 0;i<n;i++){
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char[] str = stickers[i].toCharArray();
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for (char cha:str){
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counts[i][cha-'a']++;
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}
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}
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HashMap<String,Integer> dp = new HashMap<>();
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dp.put("",0);
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int ans = process3(counts,target,dp);
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return ans == Integer.MAX_VALUE?-1:ans;
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}
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private static int process3(int[][] stickers, String t, HashMap<String, Integer> dp) {
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if (dp.containsKey(t)){
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return dp.get(t);
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}
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char[] target = t.toCharArray();
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int[] tcounts = new int[26];
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for (char cha:target){
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tcounts[cha-'a']++;
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}
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int n=stickers.length;
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int min = Integer.MAX_VALUE;
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for (int i=0;i<n;i++){
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int[] sticker = stickers[i];
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if (sticker[target[0]-'a']>0){
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StringBuilder builder = new StringBuilder();
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for (int j= 0 ;j<26;j++){
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if (tcounts[j]>0){
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int nums= tcounts[j]-sticker[j];
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for (int k=0;k<nums;k++){
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builder.append((char)(j+'a'));
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}
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}
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}
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String rest = builder.toString();
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min = Math.min(min,process3(stickers,rest,dp));
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}
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}
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int ans = min+(min==Integer.MAX_VALUE?0:1);
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dp.put(t,ans);
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return ans;
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}
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}
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package com.learn.learn.tixi.twentieth;
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public class BagMaxValue {
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public static void main(String[] args) {
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int [] weight = {3,2,4,7,3,1,7};
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int [] value = {5,6,3,19,12,4,2};
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int bag = 15;
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System.out.println(maxValue(weight,value,bag));
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System.out.println(dp(weight,value,bag));
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}
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//====================================暴力递归
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public static int maxValue(int[] w,int[] v,int bag){
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if (w==null||w.length==0||v==null||v.length==0){
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return 0;
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}
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return process(w,v,0,bag);
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}
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/*
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要与不要都测试
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*/
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public static int process(int[] w,int[] v,int index,int bag){
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if (bag<0){
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// 设置无效解
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return -1;
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}
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if (index == w.length-1){
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return 0;
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}
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// 有货，bag有空间
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// 不要当前货
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int p1 = process(w,v,index+1,bag);
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int next = process(w, v, index + 1, bag - w[index]);
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int p2 =0 ;
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if (next!=-1){
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p2 = v[index]+next;
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}
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return Math.max(p1,p2);
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}
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//========================================缓存
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// 尝试策略就是动态规划转移方程。
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// 动态规划转移方程就是尝试策略。
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public static int dp(int[] w,int[] v,int bag){
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if (w==null||w.length==0||v==null||v.length==0){
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return 0;
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}
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int n = w.length;
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int [][] dp = new int[n+1][bag+1];
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for (int index = n-1;index>=0;index--){
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for (int rest = 0;rest<=bag;rest++){
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int p1 = dp[index+1][rest];
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int p2=0;
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int next = rest-w[index]<0?-1:dp[index+1][rest-w[index]];
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if (next!=-1){
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p2 = v[index]+next;
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}
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dp[index][rest]=Math.max(p1,p2);
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}
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}
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return dp[0][bag];
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}
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}
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package com.learn.learn.tixi.twentieth;
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// https://leetcode.cn/problems/longest-common-subsequence/
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public class LongestCommonSubsequence {
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public static int longestCommonSubsequence(String s1,String s2){
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if (s1==null||s2==null||s1.length()==0||s2.length()==0){
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return 0;
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}
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char[] str1 = s1.toCharArray();
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char[] str2 = s2.toCharArray();
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return process1(str1,str2,str1.length-1,str2.length-1);
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}
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public static int process1(char[] str1,char[] str2,int i,int j){
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if (i==0&&j==0){
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return str1[i]==str2[j]?1:0;
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}else if (i==0){
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if (str2[j]==str1[i]){
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return 1;
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}else {
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process1(str1,str2,i,j-1);
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}
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}else if (j==0){
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if (str2[j]==str1[i]){
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return 1;
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}else {
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process1(str1,str2,i-1,j);
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}
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}else {
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// 样本对应，结尾该如何组织可能性
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// 不以i结尾可能以j结尾，
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int p1 = process1(str1,str2,i-1,j);
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// 即以i结尾也不以j结尾，
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int p2 = process1(str1,str2,i,j-1);
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// 即以i结尾又以j结尾
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int p3 = str1[i]==str2[j]?(1+process1(str1,str2,i-1,j-1)):0;
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return Math.max(p1,Math.max(p2,p3));
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}
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return 0;
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}
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public static int longestCommonSubsequence2(String s1,String s2){
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if (s1==null||s2==null||s1.length()==0||s2.length()==0){
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return 0;
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}
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char[] str1 = s1.toCharArray();
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char[] str2 = s2.toCharArray();
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int n= str1.length;
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int m = str2.length;
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int [][] dp = new int[n][m];
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dp[0][0]=str1[0]==str2[0]?1:0;
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for (int j=1;j<m;j++){
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dp[0][j]=str1[0]==str2[j]?1:dp[0][j-1];
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}
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for (int i=1;i<n;i++){
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dp[i][0]=str1[i]==str2[0]?1:dp[i-1][0];
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}
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for (int i=1;i<n;i++){
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for (int j=1;j<m;j++){
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int p1 = dp[i-1][j];
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// 即以i结尾也不以j结尾，
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int p2 = dp[i][j-1];
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// 即以i结尾又以j结尾
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int p3 = str1[i]==str2[j]?(1+dp[i-1][j-1]):0;
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dp[i][j]= Math.max(p1,Math.max(p2,p3));
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}
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}
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return dp[n-1][m-1];
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}
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}
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package com.learn.learn.tixi.twentieth;
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import org.apache.tomcat.util.buf.StringUtils;
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public class NumberConvert {
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public static void main(String[] args) {
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System.out.println(number("111"));
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System.out.println(dp("111"));
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System.out.println(number("305"));
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System.out.println(dp("305"));
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}
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public static int number(String str){
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if (str==null||str.length()==0){
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return 0;
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}
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return process(str.toCharArray(),0);
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}
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/*
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从0开始看有多少中转换方式
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*/
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private static int process(char[] str, int i) {
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if (i==str.length){
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// 之前的转换方式
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return 1;
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}
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if (str[i]=='0'){//之前决定有问题
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return 0;
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}
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// i 位置单转
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int ways = process(str,i+1);
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if(i+1<str.length&&(str[i]-'0')*10+str[i+1]-'0'<27){
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ways +=process(str,i+2);
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}
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return ways;
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}
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//===================================缓存
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public static int dp(String str){
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if (str==null||str.length()==0){
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return 0;
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}
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char[] strChar = str.toCharArray();
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int n = strChar.length;
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int[] dp = new int[n+1];
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// 依赖后面的位置，从右往左填写
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dp[n]=1;
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for (int i=n-1;i>=0;i--){
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if (strChar[i]!='0'){
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int ways=dp[i+1];
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if (i + 1 < strChar.length &&(strChar[i]-'0')*10+strChar[i+1]-'0'<27) {
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ways+=dp[i+2];
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}
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dp[i]=ways;
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}
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}
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return dp[0];
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}
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}
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@ -0,0 +1,27 @@
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### 背包问题
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重量数组w[]
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价值数据v[]
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bag maxvalue int
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自由挑选货物，让背包里装的货物最多
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### 数字对应
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规定1对应A，2对应B，3对应C，·····，26对应Z，
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那么一个数字字符串比如‘111’就可以转换为‘AAA’，‘KA’和‘AK’
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给定一个只有数字字符串组成的字符串str，返回对少中转换结果
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### 数组转字符串
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给定一个字符串str，给定一个字符串类型的数组arr，
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出现的字符都是小写英文arr每一个字符串，代表一张贴纸，
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你可以把当个字符剪开使用，目的是拼出来str来返回需要至少
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多少贴纸可以完成任务。
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例子：str="babac"，arr={"ba","c","abcd"}
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至少需要两张贴纸"ba"和"abcd",因为使用这两张贴纸，
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把每一个字符串单独剪开，含有2个a、2个b、1个c。是可以
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拼出str的。所以返回2
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### 最长公共子序列
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样本对应模型：
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### 模型
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1. 从左往右尝试
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2. 范围尝试
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3. 样本对应模型
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4. 业务限制模型
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