Minimum Path Sum

时间：2014-10-06 19:58:10 收藏：0 阅读：263

Given a m x n grid filled with non-negative numbers, find a path from top left to bottom right which minimizes the sum of all numbers along its path.

Note: You can only move either down or right at any point in time.

格子取数问题，另dp[i][j]表示走到(i,j)格子时，取得的最小数字，那么有

dp[i][j] = grid[i][j] + min(dp[i-1][j], dp[i][j-1])

当i=0时，只能从左边过来，当j=0时，只能从上面过来

 1 class Solution {
 2 public:
 3     int minPathSum(vector<vector<int> > &grid) {
 4         int m = grid.size();
 5         int n = grid[0].size();
 6         if( m<0 || n<0 ) return 0;
 7         vector< vector<int> > dp(grid);
 8         for(int i=1; i<n; ++i) dp[0][i] += dp[0][i-1];
 9         for(int i=1; i<m; ++i) dp[i][0] += dp[i-1][0];
10         for(int i=1; i<m; ++i)
11             for(int j=1; j<n; ++j)
12                 dp[i][j] += min(dp[i-1][j], dp[i][j-1]);
13         return dp[m-1][n-1];
14     }
15 };

原文：http://www.cnblogs.com/bugfly/p/4008593.html