120. Triangle

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

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：明显是动态规划，逐步求解。O(n)空间复杂度的方法是从下往上求。

http://www.cnblogs.com/grandyang/p/4286274.html