DP-Triangle

题目描述：

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, where n is the total number of rows in the triangle.

思路：

记f(i, j)为以(x, j)为根的最短路径和。

状态转移方程：f(i, j) = min{f(i+1， j), f(i+1, j+1)} + (i, j)。

实现：

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

DP-Triangle

原文：http://www.cnblogs.com/gatsbydhn/p/4970328.html