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.

//f(i,j) = min (f(i+1,j),f(i+1,j+1)) +a[i][j] ---->自底向上

从底端与之比较的为累加值，与相邻的元素值，在于它比较！！

  //f(i,j) = min (f(i+1,j),f(i+1,j+1)) +a[i][j] ---->自底向上    int minimumTotal(vector<vector<int> > &triangle) {                int row = triangle.size();            for(int i = row-2 ;i >=0; i--)//第一行开始        {            for(int j = 0 ; j <= i ;j++ )            {               triangle[i][j] = min(triangle[i+1][j],triangle[i+1][j+1]) + triangle[i][j];//被覆盖！            }        }        return triangle[0][0];    }

//自顶向下的方式。（稍微麻烦点，因为要考虑俩短情况）以下f[][]可以用一维数组改进。

int minimumTotal(vector<vector<int> > &triangle) {                int row = triangle.size();        int f[row][triangle[row-1].size()];        int val1,val2;        int min_val = INT_MAX;        f[0][0]=triangle[0][0];                         for(int i=1;i<row;i++)        {            f[i][i+1] = INT_MAX;           for(int j=0;j<=i;j++)            {                              val1 = j>=1?f[i-1][j-1]:INT_MAX;               val2 = j<i? f[i-1][j]:INT_MAX;               f[i][j]=min(val1,val2)+triangle[i][j];           }        }                for(int i=0;i<row;i++)        {            min_val = min(f[row-1][i],min_val);        }        return min_val;    }