标签：style blog color strong for io
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.
思路：每层的每一个节点的最短路径只有上一层的两个节点决定。从下往上计算。
1 class Solution { 2 public: 3 int minimumTotal(vector<vector<int> > &triangle) { 4 int len=triangle.size(); 5 vector<int> pathsum; 6 for(int i=0;i<triangle[len-1].size();i++) 7 pathsum.push_back(triangle[len-1][i]); 8 for(int i=len-2;i>=0;i--) 9 { 10 for(int j=0;j<triangle[i].size();j++) 11 { 12 int sum=min(pathsum[j]+triangle[i][j], pathsum[j+1]+triangle[i][j]); 13 pathsum[j]=sum; 14 } 15 } 16 return pathsum[0]; 17 } 18 };
标签：style blog color strong for io
原文地址：http://www.cnblogs.com/hicandyman/p/3829617.html
