120.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：其实由上到下和由下到上按照规则计算出来的和都是符合要求的。我们采取由下到上的顺序，初始化temp[]数组为triangle数组的最后一行。有以下状态转移方程成立：

temp[j]=min(temp[j],temp[j+1])+triangle[i][j] {i=row-2 -> 0}, 第i行计算出来的temp数组temp[j]表明从三角形最下面一行到三角形(i,j)位置的最小和。当计算到i=0时，temp[0]就是我们需要的最小和的值。

 

 
class Solution {
public:
    int minimumTotal(vector<vector<int>>& triangle) {
        int row=triangle.size();
        if(row==0)
            return 0;
        vector<int > temp(triangle[row-1].begin(),triangle[row-1].end());
        for(int i=row-2;i>-1;i--){
            int col=triangle[i].size();
            for(int j=0;j<col;j++){
                temp[j]=min(temp[j],temp[j+1])+triangle[i][j];
            }
        }
        return temp[0];
    }
};