(一)题目描述
(二)思想方法
https://blog.csdn.net/qq874455953/article/details/82806030
(三)代码实现
class Solution {
public:
int minimumTotal(vector<vector<int>>& triangle)
{ vector<vector<int>> minPath = triangle;
int n = triangle.size();
for (int i = 1; i < n; i++) {
for (int j = 0; j <= i; j++) {
//当计算节点是头结点或者尾节点时,走到此节点的路径只有一个
if (j == 0) {
minPath[i][j] = minPath[i - 1][j] + triangle[i][j];
} else if (j == i) {
minPath[i][j] = minPath[i - 1][j - 1] + triangle[i][j];
} else {
//当非头尾节点的时候, 走到节点的方式有两种
minPath[i][j] = min(minPath[i-1][j-1] + triangle[i][j], minPath[i-1][j] + triangle[i][j]);
}
}
}
//返回最后一层节点的最小路径的最小值 即目标值
return *min_element(minPath[n-1].begin(), minPath[n-1].end());
}
};