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