LeetCode -- Unique Paths II

Question:

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.

Analysis:

根据“Unique Paths”题目：

现在想象假设在grid中存在一些障碍，那么应该会有几条不同的路径呢？

1表示此处是障碍，不能走；0表示此处可以行走。

思路：根据前面两个题目的经验，仍然使用动态规划，额外申请一块空间，当此处在原网格中是1，表示此处走不通，，所以前面累计的路径个数清零。注意一些特殊情况，如start位置或finish位置为1则一定没有路径。

Answer：

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