Unique Paths II leetcode

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.

在第i个位置上设置一个障碍物后，说明位置i到最后一个格子这些路都没法走 为0

所以说明，在初始条件时，如果一旦遇到障碍物，障碍物后面所有格子的走法都是0

再看求解过程，当然按照上一题的分析dp[i][j] = dp[i-1][j] + dp[i][j-1] 的递推式依然成立.碰到了障碍物，那么这时的到这里的走法应该设为0，因为机器人只能向下走或者向右走，所以到这个点就无法通过。

时间O(m*n) 空间O(m*n) 

public class Solution {
public class Solution {
    public int uniquePathsWithObstacles(int[][] obstacleGrid) {
        if (obstacleGrid == null || obstacleGrid.length == 0 || obstacleGrid[0].length == 0) {
            return 0;
        }
        int m = obstacleGrid.length;
        int n = obstacleGrid[0].length;
        int [][] sum = new int [m][n];
        for (int i = 0; i < m; i++) {
            if (obstacleGrid[i][0] != 1) {
                sum[i][0] = 1;
            } else {
                break;//后面所有的都无法到达所以break
            }
        }
        for (int j = 0; j < n; j++) {
            if (obstacleGrid[0][j] != 1) {
                sum[0][j] = 1;
            } else {
                break;
            }
        }
        for (int i = 1; i < m; i++) {
            for (int j = 1; j < n; j++) {
                if (obstacleGrid[i][j] != 1) {
                    sum[i][j] = sum[i-1][j] + sum[i][j-1];
                } else {
                    sum[i][j] = 0;
                }
            }
        }
        return sum[m-1][n-1];
        
    }
}