leetcode——63. 不同路径 II

class Solution:
    def uniquePathsWithObstacles(self, obstacleGrid) -> int:
        if len(obstacleGrid)<1:
            return 0
        m=len(obstacleGrid)
        n=len(obstacleGrid[0])
        if m==1 and n==1:
            if obstacleGrid[0][0]==0:
                return 1
            else:
                return 0
        memo=[[0 for i in range(n)] for i in range(m)]#定义空二维数组
        for i in range(n):
            if obstacleGrid[0][i]==1:
                memo[0][i]=0
                for k in range(i+1,n):
                    memo[0][k]=0
                break
            else:
                memo[0][i]=1
        for j in range(m):
            if obstacleGrid[j][0]==1:
                memo[j:][0]=[0]*(m-j-1)
                break
            else:
                memo[j][0]=1
        for i in range(1,m):
            for j in range(1,n):
                if obstacleGrid[i][j]==1:
                    memo[i][j]=0
                else:
                    memo[i][j]=memo[i-1][j]+memo[i][j-1]
        return memo[m-1][n-1]

执行用时 :64 ms, 在所有 python3 提交中击败了76.80%的用户

内存消耗 :13.7 MB, 在所有 python3 提交中击败了5.19%的用户

 

                                                  ——2019.10.14

 

public int uniquePathsWithObstacles(int[][] obstacleGrid) {
        int m = obstacleGrid.length;
        int n = obstacleGrid[0].length;
        int[][] dp = new int[m][n];
        int i= 0;
        while(i<m){
            if(obstacleGrid[i][0] == 1){
                dp[i++][0] = 0;
                while(i<m){
                    dp[i++][0] = 0;
                }
            }
            if(i<m) {
                dp[i++][0] = 1;
            }
        }
        int j = 0;
        while(j<n){
            if(obstacleGrid[0][j] == 1){
                dp[0][j++] = 0;
                while(j<n){
                    dp[0][j++] = 0;
                }
            }
            if(j<n) {
                dp[0][j++] = 1;
            }
        }

        for(int i1 = 1;i1<m;i1++){
            for(int j1 = 1;j1<n;j1++){
                if(obstacleGrid[i1][j1] == 1){
                    dp[i1][j1] = 0;
                }else {
                    dp[i1][j1] = dp[i1 - 1][j1] + dp[i1][j1 - 1];
                }
            }
        }
        return dp[m-1][n-1];
    }

 没什么难度。

——2020.6.28