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.

思路：

经典的DP。因为限制用O（n）的空间，只需保存最下面一行的数字，然后逐行相加即可。

代码：

    int min(int a, int b){
        if(a < b)
            return a;
        return b;
    }
    int minimumTotal(vector<vector<int> > &triangle) {
        // IMPORTANT: Please reset any member data you declared, as
        // the same Solution instance will be reused for each test case.
        int n = triangle.size();
        if(n == 0)
            return 0;
        vector<int> v = triangle[n-1];
        int m = n;
        while(m > 1){
            for(int i = 0; i < m-1; i++){
                v[i] = triangle[m-2][i] + min(v[i], v[i+1]);
            }
            m--;
        }
        return v[0];
    }