代码随想录算法训练营第四十三天 | LeetCode1049. 最后一块石头的重量 II 、494. 目标和、474.一和零
一、1049. 最后一块石头的重量 II
解题代码C++:
class Solution {
public:int lastStoneWeightII(vector<int>& stones) {vector<int> dp(15001, 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++) { // 遍历物品for (int j = target; j >= stones[i]; j--) { // 遍历背包dp[j] = max(dp[j], dp[j - stones[i]] + stones[i]);}}return sum - dp[target] - dp[target];}
};
题目链接/文章讲解/视频讲解:
https://programmercarl.com/1049.%E6%9C%80%E5%90%8E%E4%B8%80%E5%9D%97%E7%9F%B3%E5%A4%B4%E7%9A%84%E9%87%8D%E9%87%8FII.html
二、494. 目标和
解题代码C++:
class Solution {
public:int findTargetSumWays(vector<int>& nums, int S) {int sum = 0;for (int i = 0; i < nums.size(); i++) sum += nums[i];if (abs(S) > sum) return 0; // 此时没有方案if ((S + sum) % 2 == 1) return 0; // 此时没有方案int bagSize = (S + sum) / 2;vector<int> dp(bagSize + 1, 0);dp[0] = 1;for (int i = 0; i < nums.size(); i++) {for (int j = bagSize; j >= nums[i]; j--) {dp[j] += dp[j - nums[i]];}}return dp[bagSize];}
};
题目链接/文章讲解/视频讲解:
https://programmercarl.com/0494.%E7%9B%AE%E6%A0%87%E5%92%8C.html
三、474.一和零
解题代码C++:
class Solution {
public:int findMaxForm(vector<string>& strs, int m, int n) {vector<vector<int>> dp(m + 1, vector<int> (n + 1, 0)); // 默认初始化0for (string str : strs) { // 遍历物品int oneNum = 0, zeroNum = 0;for (char c : str) {if (c == '0') zeroNum++;else oneNum++;}for (int i = m; i >= zeroNum; i--) { // 遍历背包容量且从后向前遍历!for (int j = n; j >= oneNum; j--) {dp[i][j] = max(dp[i][j], dp[i - zeroNum][j - oneNum] + 1);}}}return dp[m][n];}
};
题目链接/文章讲解/视频讲解:
https://programmercarl.com/0474.%E4%B8%80%E5%92%8C%E9%9B%B6.html