[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.

本题难度Medium。

一维数组

【复杂度】

时间 O(N^2) 空间 O(N)

【思路】

与Unique Paths的一维数组法一样，只不过多了一个障碍，因此对于

j!=0

的

dp[j]

赋值：

dp[j]=(obstacleGrid[i][j]==1)?0:dp[j]+dp[j-1];

但是要注意的情况是

dp[0]

的赋值，比如：

[
[0,0,0],
[1,0,0],
[0,0,0]
]

第二行第一列的值为1，那么不仅它的本行

dp[0]=0

，而且对于第三行第一列

dp[0]=0

。所以在此，对于

dp[0]

值的赋值为：

dp[0]=(dp[0]==0)?0:(obstacleGrid[i][0]==1?0:1);

【代码】

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