问题描述：

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, wheren is the total number of rows in the triangle.

基本思想：

构造一个与给定三角形同样形状的三角形，每个节点记录从根到这个节点的最小路径。

计算每个节点的从根最小距离只依赖于上面节点的状况:

1. 节点是最左节点， f[i][0] = f[i-1][0] + triangle[i][0] (i >=1)
2. 节点是最右节点， f[i][j]  = f[i-1][j-1] +triangle[i][j]
3. 节点是中间节点， f[i][j]  = min(f[i-1][j-1],f[i-1][j]) + triangle[i][j]
   

由此可以依次计算出每个节点的最小路径。

如果记录整个三角形需要O（n^2）的空间。 但是由于每行的计算只依赖于上一行的内容，所以空间复杂度可以只用O（n）。

代码：

    int minimumTotal(vector<vector<int> > &triangle) { // c++
        int rows = triangle.size();
        //check null
        if(rows == 0)
            return 0;
        
        vector<int> record(rows,triangle[0][0]);
        int prej;   //use prej to record  record[j-1] before update
                    
        //tmp remember record[j] before update;
        for(int i = 1; i < triangle.size(); i++)
            for(int j = 0; j < triangle[i].size(); j++){
                if(j == 0){
                    prej = record[j];
                    record[j] += triangle[i][j];  
                }
                // record[j] = record[j-1] + triangle[i][j];
                else if(j == triangle[i].size()-1){
                    int tmp = record[j];  
                    record[j] = prej + triangle[i][j];
                    prej = tmp;
                }
                // record[j] = ((record[j] < record[j-1])? record[j] : record[j-1]) + triangle[i][j];
                else{ 
                    int tmp = record[j];
                    record[j] = ((record[j] < prej)? record[j] : prej) + triangle[i][j];
                    prej = tmp;
                }
            }
        
        //find minmum
        int min = record[0];
        for(int i = 1; i < record.size(); i++)
            if(record[i] < min)
                min = record[i];
        return min;
    }

[leetcode]Triangle