Triangle DP求最大和

Triangle

 

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.

class Solution {public:    int minimumTotal(vector<vector<int> > &triangle) {                int n=triangle.size();        vector<int> dp(n);        for(int i=0;i<n;i++)            dp[i]=triangle[n-1][i];        for(int i=n-2;i>=0;i--)        {            for(int j=0;j<=i;j++)            {                dp[j]=triangle[i][j]+min(dp[j],dp[j+1]);            }        }        return dp[0];    }};