**一. 题目描述** Given a triangle, find the minimum path sum from top to bottom. Each step you may move to adjacent numbers on the row below. For example, given the following triangle ~~~ [ [2], [3,4], [6,5,7], [4,1,8,3] ] ~~~ The minimum path sum from top to bottom is 11 (i.e., 2 + 3 + 5 + 1 = 11). Note: Bonus point if you are able to do this using only O(n) extra space, where n is the total number of rows in the triangle. **二. 题目分析** 使用动态规划来完成。设从顶部到第`i`层的第`k`个顶点的最小路径长度表示为`f(i, k)`，则`f(i, k) = min{f(i-1,k), f(i-1,k-1)} + d(i, k)`，其中`d(i, k)`表示原来三角形数组里的第i行第k列的元素。则可以求得从第一行到最终到第`length-1`行第`k`个元素的最小路径长度，最后再比较第`length-1`行中所有元素的路径长度大小，求得最小值。 这里需要注意边界条件，即每一行中的第一和最后一个元素在上一行中只有一个邻居。而其他中间的元素在上一行中都有两个相邻元素。 **三. 示例代码** ~~~ class Solution { public: int minimumTotal(vector<vector<int> > &triangle) { vector< vector<int> >::size_type length = triangle.size(); if(length == 0){ return 0; } int i, j; for(i=1;i<length;i++){ vector<int>::size_type length_inner = triangle[i].size(); for(j=0;j<length_inner;j++){ if(j == 0){ triangle[i][j] = triangle[i][j] + triangle[i-1][j]; } else if(j == length_inner - 1){ triangle[i][j] = triangle[i][j] + triangle[i-1][j-1]; } else{ triangle[i][j] = (triangle[i][j] + triangle[i-1][j-1] < triangle[i][j] + triangle[i-1][j] ? triangle[i][j] + triangle[i-1][j-1]:triangle[i][j] + triangle[i-1][j]); } } } int min_path = triangle[length-1][0]; for(i=1;i<triangle[length-1].size();i++){ min_path = (min_path < triangle[length-1][i]?min_path:triangle[length-1][i]); } return min_path; } }; ~~~ **四. 小结** 无