leetcode----63. Unique Paths II

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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.

版本一

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

版本2

public int uniquePathsWithObstacles(int[][] obstacleGrid) {
        if(obstacleGrid == null || obstacleGrid.length==0 || obstacleGrid[0].length==0)  
            return 0; 
        int m = obstacleGrid.length, n = obstacleGrid[0].length;
        int [] dp=new int[n];
        dp[0]=1;
        for(int i = 0; i < m; i++){
            for(int j = 0; j < n; j++){
                if(obstacleGrid[i][j]==1){
                    dp[j]=0;
                }
                else if(j>0){
                    dp[j]+=dp[j-1];
                }
            }
        }
        return dp[n-1];
    }