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前言

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有N件物品和一个最多能背重量为W的背包。第i件物品的重量是weight[i]，得到的价值是value[i] 。每件物品都有无限个（也就是可以放入背包多次），求解将哪些物品装入背包里物品价值总和最大。

一、力扣416. 分割等和子集

class Solution {public boolean canPartition(int[] nums) {int sum = 0;for(int a : nums){sum += a;}if(sum % 2 == 1){return false;}sum /= 2;int n = nums.length;int[] dp = new int[sum+1];for(int i = 0; i <= sum; i ++){if(nums[0] <= i){dp[i] = nums[0];}}for(int i = 1; i < n; i ++){for(int j = sum; j >= nums[i]; j --){dp[j] = Math.max(dp[j],dp[j-nums[i]] + nums[i]);}}if(dp[sum] == sum){return true;}return false;}
}

二、力扣1049. 最后一块石头的重量 II

class Solution {public int lastStoneWeightII(int[] stones) {int sum = 0;for(int a : stones){sum += a;}int count = sum;sum /= 2;int[] dp = new int[sum+1];for(int i = 0; i < stones.length; i ++){for(int j = sum; j >= stones[i]; j --){dp[j] = Math.max(dp[j], dp[j-stones[i]] + stones[i]);}}return count - (2*dp[sum]);}
}

三、力扣494. 目标和

在这里插入代码片class Solution {public int findTargetSumWays(int[] nums, int target) {int sum = 0;for(int a : nums){sum += a;}if((target+sum)%2 != 0){return 0;}if(Math.abs(target) > sum){return 0;}int A = (target+sum)/2;A = A > 0?A:-A;int[] dp = new int[A+1];dp[0] = 1;for(int i = 0; i < nums.length; i ++){for(int j = A; j >= nums[i]; j --){dp[j] += dp[j-nums[i]];}}return dp[A];}
}

四、力扣474. 一和零

class Solution {public int findMaxForm(String[] strs, int m, int n) {int[][] dp = new int[m+1][n+1];for(int i = 0; i < strs.length; i ++){int oneSize = 0, zeroSize = 0;for(char c : strs[i].toCharArray()){if(c == '0'){zeroSize ++;}else{oneSize ++;}}for(int j = m; j >=zeroSize; j -- ){for(int k = n; k >= oneSize; k --){dp[j][k] = Math.max(dp[j][k],dp[j-zeroSize][k-oneSize]+1);}}}return dp[m][n];}
}