leetcode------Unique Paths II

标题：	Unique Paths II
通过率：	28%
难度：	中等

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.

与第一个版本一样，但是原始数组中会出现1的情况那么针对1来处理，如果遇到1则将其置-1，然后在路径相加中如果是-1则不进行相加

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         boolean flag=true;

 6         if(obstacleGrid[0][0]==1)return 0;

 7         for(int i=0;i<m;i++){

 8             if(obstacleGrid[i][0]!=1&&flag)

 9                 obstacleGrid[i][0]=1;

10             else {

11                 obstacleGrid[i][0]=-1;

12                 flag=false;

13             }

14         }

15         flag=true;

16         for(int i=1;i<n;i++){

17             if(obstacleGrid[0][i]!=1&&flag)

18                 obstacleGrid[0][i]=1;

19             else {

20                 obstacleGrid[0][i]=-1;

21                 flag=false;

22             }

23         }

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

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

26                 if(obstacleGrid[i][j]!=1){

27                     if(obstacleGrid[i-1][j]==-1&&obstacleGrid[i][j-1]==-1)

28                         obstacleGrid[i][j]=0;

29                     if(obstacleGrid[i-1][j]==-1)

30                         obstacleGrid[i][j]+=obstacleGrid[i][j-1];

31                     else if(obstacleGrid[i][j-1]==-1)

32                         obstacleGrid[i][j]+=obstacleGrid[i-1][j];

33                     else obstacleGrid[i][j]=obstacleGrid[i-1][j]+obstacleGrid[i][j-1];

34                 }

35                 else{

36                     obstacleGrid[i][j]=-1;

37                 }

38             }

39         }

40             if(obstacleGrid[m-1][n-1]==-1)return 0;

41             else 

42                 return obstacleGrid[m-1][n-1];

43         

44     }

45 }