**一. 題目描述** 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. 題目分析** 使用動態規劃來完成。設從頂部到第`i`層的第`k`個頂點的最小路徑長度表示為`f(i, k)`，則`f(i, k) = min{f(i-1,k), f(i-1,k-1)} + d(i, k)`，其中`d(i, k)`表示原來三角形數組里的第i行第k列的元素。則可以求得從第一行到最終到第`length-1`行第`k`個元素的最小路徑長度，最后再比較第`length-1`行中所有元素的路徑長度大小，求得最小值。 這里需要注意邊界條件，即每一行中的第一和最后一個元素在上一行中只有一個鄰居。而其他中間的元素在上一行中都有兩個相鄰元素。 **三. 示例代碼** ~~~ 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. 小結** 無