Leetcode: 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.

与Unique Path, Minimum Path Sum题目解法类似，都是DP问题。建立一个新的矩阵，矩阵每个元素记录的是从upper left到该点的路径数。与Unique Path不同的是：写递归式的时候，能不能代入下一层递归需要检验obstacleGrid里该点是不是一个obstacle，如果是，该层返回0且不继续递归

Iteration方法：21-26行可删，因为如果那里是obstacle，那里在res矩阵里也是0，加它也无妨

 1 public class Solution {

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

 3         int m = obstacleGrid.length;

 4         int n = obstacleGrid[0].length;

 5         int[][] res = new int[m][n];

 6         for (int k=0; k<n; k++) {

 7             if (obstacleGrid[0][k] != 1) {

 8                 res[0][k] = 1;

 9             }

10             else break;

11         }

12         for (int t=0; t<m; t++) {

13             if (obstacleGrid[t][0] != 1) {

14                 res[t][0] = 1;

15             }

16             else break;

17         }

18         for (int i=1; i<m; i++) {

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

20                 if (obstacleGrid[i][j] == 1) continue;

21                 else {

22                     res[i][j] = res[i-1][j] + res[i][j-1];

23                 }

24             }

25         }

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

27     }

28 }