【leetcode刷题笔记】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.

Note: m and n will be at most 100.

---

 

题解：和http://www.cnblogs.com/sunshineatnoon/p/3798167.html非常相似，要注意两点：

1. 有障碍的地方能走到的方法数目是0；
2. 初始化结果矩阵的时候，如果第一行（列）的某个位置有障碍，那么这一行（列）该元素后面所有的元素都是0，不能置为1；比如给定障碍矩阵[1,0,0]，那么初始化以后的矩阵应该为[0,0,0]，而不是[0,1,1]，因为(0,0)处有障碍，那么该矩阵的任何位置都是无法到达的。

代码如下：

 1 public class Solution {

 2     public int uniquePathsWithObstacles(int[][] obstacleGrid) {

 3         int m = obstacleGrid.length;

 4         int n = obstacleGrid[0].length;

 5         if(m==0 && n == 0)

 6             return 0;

 7         

 8         int[][] PathNum = new int[m][n];

 9         for(int i = 0;i < m;i++)

10             if(obstacleGrid[i][0] != 1)

11                 PathNum[i][0] = 1;

12             else 

13                 break;

14         for(int i = 0;i < n;i++)

15             if(obstacleGrid[0][i] != 1)

16                 PathNum[0][i] = 1;

17             else

18                 break;

19         

20         for(int i = 1;i < m;i++)

21             for(int j = 1;j < n;j++){

22                 if(obstacleGrid[i][j] != 1)

23                     PathNum[i][j] = PathNum[i-1][j]+PathNum[i][j-1]; 

24             }

25         

26         return PathNum[m-1][n-1];

27     }

28 }