leetcode打卡#day44 1049. 最后一块石头的重量 II、494. 目标和、474. 一和零
1049. 最后一块石头的重量 II
class Solution {
public:
int lastStoneWeightII(vector& stones) {
vector 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[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& nums, int target) {
//分成+、-法2个集合, left: +的集合, right: -的集合
//left = (sum + target) / 2
int 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 dp(left+1, 0); //装满容器有dp[i]种方法
dp[0] = 1;
for (int i = 0; i = nums[i]; j--) {
dp[j] += dp[j - nums[i]];
}
}
return dp[left];
}
};
474. 一和零
class Solution {
public:
int findMaxForm(vector& strs, int m, int n) {
//最多有m个0, n个1
vector<vector> dp(m+1, vector(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--) { // 0
for (int j = n; j >= y; j--) {// 1
dp[i][j] = max (dp[i][j], dp[i-x][j-y] + 1);
}
}
}
return dp[m][n];
}
};