标签:leetcode
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.
解题思路:动态规划比较简单的题。用一个二维数组进行存到某个位置的最小值,用两个for i=0~m-1,for j=0~n-1,进行搜索,对两种情况一个j+1,与 i+1,如果以前的存在大于现在的,特换,如果以前值没有,那就存入。思路比较简单。
class Solution { public: int minPathSum(vector<vector<int> > &grid) { size_t m = grid.size(); size_t n = grid[0].size(); int temp[m][n]; for (size_t i = 0; i != m; i++) for(size_t j = 0; j != n; j++) temp[i][j] = -1; temp[0][0] = grid[0][0]; for (size_t i = 0; i != m; i++) { for(size_t j = 0; j != n; j++) { if(j != n-1) { if(temp[i][j+1] != -1) temp[i][j+1] = (temp[i][j+1] > temp[i][j] + grid[i][j+1] ? temp[i][j] + grid[i][j+1] : temp[i][j+1]); else temp[i][j+1] =temp[i][j] + grid[i][j+1]; } if(i != m-1) { if(temp[i+1][j] != -1) temp[i+1][j] = (temp[i+1][j] > temp[i][j] + grid[i+1][j] ? temp[i][j] + grid[i+1][j] : temp[i+1][j]); else temp[i+1][j] = temp[i][j] + grid[i+1][j]; } } } return temp[m-1][n-1]; } };
标签:leetcode
原文地址:http://blog.csdn.net/vanish_dust/article/details/42720291