dp4

1049. 最后一块石头的重量 II

class Solution {
public:int lastStoneWeightII(vector<int>& stones) {int n = stones.size();int s = reduce(stones.begin(), stones.end());int t = s / 2;vector<vector<int>> dp(n + 1, vector<int>(t + 1));dp[0][0] = 0;for(int i = 0; i < n; i++){int x = stones[i];for(int j = 0; j <= t; j++){dp[i + 1][j] = dp[i][j];if(j >= stones[i])dp[i + 1][j] = max(dp[i + 1][j], dp[i][j - x] + x);}}return s - dp[n][t] - dp[n][t];}
};

494. 目标和

class Solution {
public:int findTargetSumWays(vector<int>& nums, int target) {int n = nums.size();target += reduce(nums.begin(), nums.end());if(target < 0|| target % 2)return 0;target = target / 2;vector<vector<int>> dp(n+1, vector<int>(target + 1));dp[0][0] = 1;for(int i = 0; i < n; i++){int x = nums[i];for(int j = 0; j <= target; j++){if(j >= x)dp[i + 1][j] = dp[i][j] + dp[i][j - x];elsedp[i + 1][j] = dp[i][j];}}return dp[n][target];}
};

474. 一和零

class Solution {
public:int findMaxForm(vector<string>& strs, int m, int n) {vector<vector<int>> dp(m + 2, vector<int>(n + 2, 0));for(string &str: strs){int x = 0, y = 0;for(char &c: str){if(c == '0')x++;if(c == '1')y++;}for(int i = m; i >= x; i--){for(int j = n; j >= y; j--)dp[i][j] = max(dp[i - x][j - y] + 1, dp[i][j]);}}return dp[m][n];}
};