标签：number class ext this pat bsp desc cto lse
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
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.
c++
class Solution { public: int dp[1005][1005]; int minimumTotal(vector<vector<int>>& triangle) { int len = triangle.size(); dp[0][0] = triangle[0][0]; for(int i=1;i<len;i++) { int l = triangle[i].size(); for(int j=0;j<l;j++) { if(j==0) dp[i][j] = triangle[i][j]+dp[i-1][j]; else if(j==l-1) dp[i][j] = triangle[i][j]+dp[i-1][j-1]; else dp[i][j] = min(triangle[i][j]+dp[i-1][j],triangle[i][j]+dp[i-1][j-1]); } } int ans = 9999999; for(int i=0;i<triangle[len-1].size();i++) { ans = min(ans,dp[len-1][i]); } return ans; }
标签：number class ext this pat bsp desc cto lse
原文地址：https://www.cnblogs.com/dacc123/p/9315898.html
