【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 is11(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.

题意：给定一个三角形，找到一条从顶部到底部的路径，要求路径上的和的值最小，每一步你只能往下一行的相邻位置走下去；

这种典型的利用动态规划方式思考的；

设f(i,j)是从第一行到第i行，第j个数字的最小和值，

那么f(i,j)要么是f(i-1, j-1) 而来，要么是f(i-1, j)而来，

因此最优子结构为

f(i,j)=min(f(i-1, j-1), f(i-1, j))+triangle[i][j]


通过这种方式，可以得到从第一行到第n行的所有点的最小和值，然后
取第n行的最小值即可

class Solution{public:int minimumTotal(vector<vector<int>> & triangle){if (triangle.size()==0){return 0;}vector<vector<int>> dp;vector<int> vec;vec.push_back(triangle[0][0]);dp.push_back(vec);vec.resize(0);for (int i=1; i<triangle.size(); i++){for (int j=0; j<triangle[i].size(); j++){if (j==0){vec.push_back(triangle[i][0]);}else if(j==triangle[i].size()-1){vec.push_back(triangle[i][j]+dp[i-1][j-1]);}else{vec.push_back(min(dp[i - 1][j - 1] + triangle[i][j], dp[i - 1][j] + triangle[i][j]));}}dp.push_back(vec);vec.resize(0);}int last = dp.size() - 1;int minpath = dp[last][0];for (int i = 1; i<dp[last].size(); i++)if (dp[last][i]<minpath)minpath = dp[last][i];return minpath;}};