标签：ati 路径 turn span numbers number output path The
算法描述：
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.
Example:
Input: [ [1,3,1], [1,5,1], [4,2,1] ] Output: 7 Explanation: Because the path 1→3→1→1→1 minimizes the sum.
解题思路：求最小路径和，用动态规划法。dp[i][j] = min(dp[i-1][j],dp[i][j-1]) + grid[i][j];
int minPathSum(vector<vector<int>>& grid) { int m = grid.size(); int n = grid[0].size(); vector<vector<int>> dp(m, vector<int>(n,0)); dp[0][0] = grid[0][0]; for(int i=1; i < m; i++) dp[i][0] = grid[i][0] + dp[i-1][0]; for(int j=1; j < n; j++) dp[0][j] = grid[0][j] + dp[0][j-1]; for(int i =1; i < m; i++){ for(int j=1; j < n; j++){ dp[i][j] = min(dp[i-1][j],dp[i][j-1])+grid[i][j]; } } return dp[m-1][n-1]; }
标签：ati 路径 turn span numbers number output path The
原文地址：https://www.cnblogs.com/nobodywang/p/10339417.html
