1049. 最后一块石头的重量 II
class Solution {
public:int lastStoneWeightII(vector<int>& stones) {vector<int> dp(1501, 0);int sum = 0;for (int i = 0; i < stones.size(); i++) {sum += stones[i];}int target = sum / 2;for (int i = 0; i < stones.size(); i++) {for (int j = target; j >= stones[i]; j--) {dp[j] = max(dp[j], dp[j - stones[i]] + stones[i]);}}return sum - dp[target] - dp[target];}
};
494. 目标和
class Solution {
public:int findTargetSumWays(vector<int>& nums, int target) {//分成+、-法2个集合, left: +的集合， right: -的集合//left = （sum + target) / 2int sum = 0;for (int i = 0; i < nums.size(); i++) {sum += nums[i];}if ((sum + target)%2 == 1) return 0;int left = (sum - target) / 2;vector<int> dp(left+1, 0); //装满容器有dp[i]种方法dp[0] = 1;for (int i = 0; i < nums.size(); i++) {for (int j = left; j >= nums[i]; j--) {dp[j] += dp[j - nums[i]];}}return dp[left];}
};
474. 一和零
class Solution {
public:int findMaxForm(vector<string>& strs, int m, int n) {//最多有m个0, n个1vector<vector<int>> dp(m+1, vector<int>(n+1, 0));for(string str : strs) {int x = 0, y = 0;for (char c : str) {//统计0和1的数量if (c == '0') x++;else y++;}for (int i = m; i >= x; i--) { // 0for (int j = n; j >= y; j--) {// 1dp[i][j] = max (dp[i][j], dp[i-x][j-y] + 1);}}}return dp[m][n];}
};
