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.

题意：接着Unique Paths http://blog.csdn.net/crazy__chen/article/details/46408253，如果地图上有障碍物，此障碍物不能通过

在数组中用1表示障碍物，求从起点到终点有多少种走法。

解法1：和Unique Paths一样，使用动态规划。如果当前格子为1，则flag[i][j]设置为0即可。

[java] view plain copy

1. public class Solution {  
2.     public int uniquePathsWithObstacles(int[][] obstacleGrid) {  
3.         int m = obstacleGrid.length;    
4.         int n = obstacleGrid[0].length;    
5.         int[][] flag = new int[m][n];    
6.         if(obstacleGrid[0][0]==1) return 0;    
7.         for(int i=m-1;i>=0;i--){    
8.             for(int j=n-1;j>=0;j--){         
9.                 if(obstacleGrid[i][j]==1){//如果是障碍物，不能通过，设为0    
10.                     flag[i][j] = 0;    
11.                 }else if(i+1==m&&j+1==n){//如果是终点，初始化为1    
12.                     flag[i][j] = 1;    
13.                 }else if(i+1==m){//下边界点    
14.                     flag[i][j] = flag[i][j+1];    
15.                 }else if(j+1==n){//右边界点    
16.                     flag[i][j] = flag[i+1][j];    
17.                 }else{//其余节点，公式计算    
18.                     flag[i][j] = flag[i+1][j]+flag[i][j+1];    
19.                 }    
20.             }    
21.         }    
22.         return flag[0][0];    
23.     }  
24. }  

原文链接http://blog.csdn.net/crazy__chen/article/details/46409243