【LeetCode】64. Minimum Path Sum

Minimum Path Sum

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：

典型的动态规划。开辟m*n的矩阵path，path[i][j]存放从首元素(grid[0][0])到当前元素(grid[i][j])的最短路径长度。

对于每个元素来说，路径是从上或者从左边来的。

也就是说path[i][j] = min(path[i-1][j]+path[i][j-1]) + grid[i][j]。

注意：初始化第一行第一列。

class Solution {
public:int minPathSum(vector<vector<int> > &grid) {if(grid.empty() || grid[0].empty())return 0;int m = grid.size();int n = grid[0].size();vector<vector<int> > path(m, vector<int>(n, 0));path[0][0] = grid[0][0];for(int i = 1; i < m; i ++)path[i][0] = path[i-1][0] + grid[i][0];for(int i = 1; i < n; i ++)path[0][i] = path[0][i-1] + grid[0][i];for(int i = 1; i < m; i ++){for(int j = 1; j < n; j ++){path[i][j] = min(path[i-1][j], path[i][j-1]) + grid[i][j];}}return path[m-1][n-1];}
};

转载于:https://www.cnblogs.com/ganganloveu/p/3948069.html

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