代码随想录day35 36

想进个大厂2026-02-05 8:31

416. 分割等和子集

cpp 复制代码

class Solution {
public:
    bool canPartition(vector<int>& nums) {
        int sum = 0;
        for (auto a : nums) {
            sum += a;
        }
        if (sum % 2 != 0)
            return false;
        sum /= 2;
        vector<int> dp(sum + 1);
        for (int i = 0; i < nums.size(); i++) {
            for (int j = sum; j >= nums[i]; j--) {
                dp[j] = max(dp[j], dp[j - nums[i]] + nums[i]);
            }
        }
        return dp[sum] == sum;
    }
};

1049. 最后一块石头的重量 II

cpp 复制代码

class Solution {
public:
    int lastStoneWeightII(vector<int>& stones) {
        int total = 0;
        for (auto a : stones) {
            total += a;
        }
        int sum = total / 2;
        vector<int> dp(sum + 1);
        for (int i = 0; i < stones.size(); i++) {
            for (int j = sum; j >= stones[i]; j--) {
                dp[j] = max(dp[j], dp[j - stones[i]] + stones[i]);
            }
        }
        return total - 2 * dp[sum];
    }
};

494. 目标和

cpp 复制代码

class Solution {
public:
    int findTargetSumWays(vector<int>& nums, int target) {
        int sum = 0;
        for (int i = 0; i < nums.size(); i++)
            sum += nums[i];
        if (abs(target) > sum)
            return 0;
        if ((target + sum) % 2 == 1)
            return 0;
        int bagSize = (target + 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];
    }
};

474. 一和零

cpp 复制代码

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

其实还是相当于一维背包，strs是遍历物品，dp $i$ $j$ 本应是一个维度，本题存在两个背包容量，分别是0和1的个数。