[Leetcode] 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.

Solution：dp

public class Solution {
    public int uniquePathsWithObstacles(int[][] obstacleGrid) {
        int M=obstacleGrid.length;
        int N=obstacleGrid[0].length;
        int[][] dp=new int[M][N];
        
        for(int i=0;i<M;++i){
            if(obstacleGrid[i][0]==1){
                dp[i][0]=0;
                break;
            }else{
                dp[i][0]=1;
            }
        
        }
        for(int i=0;i<N;++i){
            if(obstacleGrid[0][i]==1){
                dp[0][i]=0;
                break;
            }else{
                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];
    }
}

以下为大神的代码：https://github.com/mengli/leetcode/blob/master/UniquePathsII.java

public class UniquePathsII {
    public int uniquePathsWithObstacles(int[][] obstacleGrid) {
        int m = obstacleGrid.length;
        if (m == 0)
            return 0;
        int n = obstacleGrid[0].length;
        int[][] map = new int[m][n];
        map[0][0] = obstacleGrid[0][0] == 1 ? 0 : 1;
        for (int i = 1; i < m; i++) {
            map[i][0] = obstacleGrid[i][0] == 1 ? 0 : map[i - 1][0];
        }
        for (int j = 1; j < n; j++) {
            map[0][j] = obstacleGrid[0][j] == 1 ? 0 : map[0][j - 1];
        }
        for (int i = 1; i < m; i++) {
            for (int j = 1; j < n; j++) {
                map[i][j] = obstacleGrid[i][j] == 1 ? 0 : map[i - 1][j]
                        + map[i][j - 1];
            }
        }
        return map[m - 1][n - 1];
    }
}