LetCode 120. 三角形最小路径和

// O(n)的额外空间
class Solution {
public:
    int minimumTotal(vector<vector<int>>& triangle) {
        vector<int> res(triangle.size(), 0);
        res[0] = triangle[0][0];
        // 从上往下，从右往左遍历
        for (int i = 1; i < triangle.size(); ++i){
            for (int j = i; j >= 0; --j){
                if (j == 0)
                    res[j] += triangle[i][j];
                else if (j == i)
                    res[j] = res[j - 1] + triangle[i][j];
                else
                    res[j] = min(res[j - 1], res[j]) + triangle[i][j];
            }
        }
        return *min_element(res.begin(),res.end());
    }
};

static int x=[](){
    std::ios::sync_with_stdio(false);
    cin.tie(NULL);
    return 0;
}();