leetcode——64.最小路径和

class Solution:
    def minPathSum(self, grid) -> int:
        m=len(grid)
        n=len(grid[0])
        if m==0 or n==0:
            return 0
        if m==1 and n==1:
            return grid[0][0]
        for i in range(1,n):
            grid[0][i]=grid[0][i]+grid[0][i-1]
        for j in range(1,m):
            grid[j][0]=grid[j-1][0]+grid[j][0]
        for i in range(1,n):
            for j in range(1,m):
                grid[j][i]=min(grid[j][i]+grid[j-1][i],grid[j][i]+grid[j][i-1])
        return grid[m-1][n-1]

执行用时 :196 ms, 在所有 python3 提交中击败了22.46%的用户

内存消耗 :15.4 MB, 在所有 python3 提交中击败了5.12%的用户

 

                                                                                       ——2019.10.14

 

时隔多日，用java再复习一遍，这道题倒是没啥难的。

public int minPathSum(int[][] grid) {
        int m = grid.length;
        int n = grid[0].length;
        int[][] dp = new int[m][n];
        dp[0][0] = grid[0][0];
        for(int j = 1;j<n;j++){
            dp[0][j] = dp[0][j-1]+grid[0][j];
        }
        for(int i = 1;i<m;i++){
            dp[i][0] = dp[i-1][0] + grid[i][0];
        }
        for(int i = 1;i<m;i++){
            for(int j = 1;j<n;j++){
                dp[i][j] = Math.min(dp[i-1][j] + grid[i][j],dp[i][j-1] + grid[i][j]);
            }
        }
        return dp[m-1][n-1];
    }

 ——2020.6.27