[leedcode 63] 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.

public class Solution {
    public int uniquePathsWithObstacles(int[][] obstacleGrid) {
        //本题和上一题的不同是增加了障碍，如何结合动态规划考虑障碍是关键
        //通过分析本题，如果该位置有障碍，那么到达该位置的路径就是0；
        //通过这个思路，只需要在计算a[i][j]时，考虑obstacleGrid[i][j]是否为1即可
        //需要注意求第一行和第一列，因为前面的障碍会使得后面的所有路径都堵塞，因此最好的办法是设置两个标记
        //flag代表列标记，flag1代表行标记
        int m=obstacleGrid.length;
        int n=obstacleGrid[0].length;
        int a[][]=new int[m][n];
        int flag=0;
        int flag1=0;
        for(int i=0;i<m;i++){
            if(flag==0&&obstacleGrid[i][0]==0){
                a[i][0]=1;
            }else{
                a[i][0]=0;
                flag=1;
            }
            for(int j=0;j<n;j++){
                if(i==0){
                    if(flag1==0&&obstacleGrid[i][j]==0)
                        a[i][j]=1;
                    else {
                        a[i][j]=0;
                        flag1=1;
                    } 

                }
            if(i>0&&j>0){
                if(obstacleGrid[i][j]==1){
                    a[i][j]=0;
                }else{
                    a[i][j]=a[i-1][j]+a[i][j-1];
                }
            }
            }
        }
        return a[m-1][n-1];
    }
}