[LeetCode]Unique Paths II

該題和之前的題目之間的區別就是某些單元是不可達的，這可以加上判斷語句即可
算法1：

public class Solution {
    public int uniquePathsWithObstacles(int[][] obstacleGrid) {
        int pathsNum = 0,m = obstacleGrid.length,n=obstacleGrid[0].length;
        if(obstacleGrid[0][0] == 1 || obstacleGrid[m-1][n-1] == 1)return 0;

        int [][] paths = new int[m][n];
        for(int i = 0; i< m; i ++){
            for(int j = 0; j < n; j ++){
                if(i > 0 && j > 0){
                    if(obstacleGrid[i][j-1] == 1 &&  obstacleGrid[i-1][j] ==0){
                        paths[i][j] = paths[i-1][j];
                    }else if(obstacleGrid[i-1][j] == 1 && obstacleGrid[i][j-1] == 0){
                        paths[i][j] = paths[i][j-1];
                    }else if(obstacleGrid[i-1][j] == 0 && obstacleGrid[i][j-1] == 0){
                        paths[i][j] = paths[i][j-1] + paths[i-1][j];
                    }else {
                        paths[i][j] = 0;
                    }
                }else if(i > 0 && j == 0){
                    if(obstacleGrid[i-1][j] == 0){
                        paths[i][j] = paths[i-1][j];
                    }else {
                        paths[i][j] = 0;
                    }

                }else if(i == 0 && j > 0){
                    if(obstacleGrid[i][j-1] == 0){
                        paths[i][j] = paths[i][j-1];
                    }else {
                        paths[i][j] = 0;
                    }
                }else {
                    if(obstacleGrid[0][0] == 1)
                        return 0;
                    else paths[0][0] = 1;
                }
            }
        }

        return paths[m-1][n-1];
    }

}

感覺自己做的太複雜了，別人的簡單版本如下：

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