Leetcode63 Unique Paths II

Posted 2022-11-25 xuweimdm

Unique Paths II

Follow up for “Unique Paths”:

Now consider if some obstacles are added to the grids. How many unique paths would there be?

An obstacle and empty space is marked as 1 and 0 respectively in the grid.

For example,
There is one obstacle in the middle of a 3x3 grid as illustrated below.

[
  [0,0,0],
  [0,1,0],
  [0,0,0]
]

The total number of unique paths is 2.

Note: m and n will be at most 100.

Solution1

• 和上一道题思路一样，只是中间添加了一些障碍物，所以只是在障碍物的地方就要将其设为0.

public class Solution 
    public int uniquePathsWithObstacles(int[][] obstacleGrid) 
        int m = obstacleGrid.length;
        if(m==0) return 0;
        int n = obstacleGrid[0].length;
        if(n==0||obstacleGrid[0][0]==1) return 0;
        int[][] dp = new int[m][n];
        dp[0][0] = 1;
        for(int i=1;i<n;i++) if(obstacleGrid[0][i]!=1) dp[0][i] += dp[0][i-1];//这里必须要从左边得到
        for(int i=1;i<m;i++)
            if(obstacleGrid[i][0]!=1) dp[i][0] = dp[i-1][0];
            for(int j=1;j<n;j++)
                if(obstacleGrid[i][j]==1) dp[i][j] = 0;
                else dp[i][j] = dp[i-1][j] + dp[i][j-1];
            
        
        return dp[m-1][n-1];       
    

Solution2

• 用滚动数组

public class Solution 
    public int uniquePathsWithObstacles(int[][] obstacleGrid) 
        int m = obstacleGrid.length;
        if(m==0) return 0;
        int n = obstacleGrid[0].length;
        if(n==0||obstacleGrid[0][0]==1) return 0;
        int[] dp = new int[n];
        dp[0] = 1;
        for(int i=1;i<n;i++) if(obstacleGrid[0][i]!=1) dp[i] = dp[i-1];
        for(int i=1;i<m;i++)
            if(obstacleGrid[i][0]==1) dp[0] = 0;
            for(int j=1;j<n;j++)
                if(obstacleGrid[i][j]==1) dp[j] = 0;
                else dp[j] += dp[j-1];
            
        
        return dp[n-1];