代码随想录：最后一块石头的重量二

背包容量一定，最多能装多少

class Solution {
public:
    int lastStoneWeightII(vector<int>& stones) {
        // 让一组石块在小于等于总重量一半的情况下尽可能的大
        int sum = 0;
        for (int stone : stones) {
            sum += stone;
        }

        int target = sum / 2;

        vector<vector<int>> dp =
            vector<vector<int>>(stones.size(), vector<int>(target + 1, 0));

        for (int j = 0; j <= target; j++) {
            if (j >= stones[0])
                dp[0][j] = stones[0];
        }

        for (int i = 1; i < stones.size(); i++) {
            for (int j = 1; j <= target; j++) {
                if (j >= stones[i]) {
                    dp[i][j] =
                        max(dp[i - 1][j], dp[i - 1][j - stones[i]] + stones[i]);
                } else {
                    dp[i][j] = dp[i - 1][j];
                }
            }
        }

        return sum - 2*dp[stones.size() - 1][target];
    }
};