[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.

Solution：dp

 1 public class Solution {
 2     public int uniquePathsWithObstacles(int[][] obstacleGrid) {
 3         int M=obstacleGrid.length;
 4         int N=obstacleGrid[0].length;
 5         int[][] dp=new int[M][N];
 6         
 7         for(int i=0;i<M;++i){
 8             if(obstacleGrid[i][0]==1){
 9                 dp[i][0]=0;
10                 break;
11             }else{
12                 dp[i][0]=1;
13             }
14         
15         }
16         for(int i=0;i<N;++i){
17             if(obstacleGrid[0][i]==1){
18                 dp[0][i]=0;
19                 break;
20             }else{
21                 dp[0][i]=1;
22             }
23         }
24         
25         for(int i=1;i<M;++i){
26             for(int j=1;j<N;++j){
27                 if(obstacleGrid[i][j]==1){
28                     dp[i][j]=0;
29                 }else{
30                     dp[i][j]=dp[i-1][j]+dp[i][j-1];
31                 }
32             }
33         }
34         
35         return dp[M-1][N-1];
36     }
37 }

以下为大神的代码：https://github.com/mengli/leetcode/blob/master/UniquePathsII.java

 1 public class UniquePathsII {
 2     public int uniquePathsWithObstacles(int[][] obstacleGrid) {
 3         int m = obstacleGrid.length;
 4         if (m == 0)
 5             return 0;
 6         int n = obstacleGrid[0].length;
 7         int[][] map = new int[m][n];
 8         map[0][0] = obstacleGrid[0][0] == 1 ? 0 : 1;
 9         for (int i = 1; i < m; i++) {
10             map[i][0] = obstacleGrid[i][0] == 1 ? 0 : map[i - 1][0];
11         }
12         for (int j = 1; j < n; j++) {
13             map[0][j] = obstacleGrid[0][j] == 1 ? 0 : map[0][j - 1];
14         }
15         for (int i = 1; i < m; i++) {
16             for (int j = 1; j < n; j++) {
17                 map[i][j] = obstacleGrid[i][j] == 1 ? 0 : map[i - 1][j]
18                         + map[i][j - 1];
19             }
20         }
21         return map[m - 1][n - 1];
22     }
23 }

 

posted @ 2014-10-24 16:27  Phoebe815  阅读(116)  评论(0)  编辑  收藏  举报