Leetcode 63. 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.

思路：动态规划题。

一开始边界的初始化很重要，其他的位置，如果是1，则置为0，否则的dp[i][j] = dp[i-1][j] + dp[i][j-1];

 

 1 class Solution {
 2 public:
 3     int uniquePathsWithObstacles(vector<vector<int>>& obstacleGrid) {
 4         if(obstacleGrid.size() == 0 || obstacleGrid[0].size() == 0)
 5             return 0;
 6         int dp[101][101], i, j;
 7         if(obstacleGrid[0][0] == 0){
 8             dp[0][0] = 1;
 9         }
10         else{
11             return 0;
12         }
13         
14         for(i = 1; i < obstacleGrid.size(); i++){
15             if(obstacleGrid[i][0] == 0 && dp[i-1][0] == 1)
16                 dp[i][0] = 1;
17             else
18                 dp[i][0] = 0;
19         }
20         
21         for(j = 1; j < obstacleGrid[0].size(); j++){
22             if(obstacleGrid[0][j] == 0 && dp[0][j-1] == 1)
23                 dp[0][j] = 1;
24             else
25                 dp[0][j] = 0;
26         }
27         
28         for( i = 1; i < obstacleGrid.size(); i++){
29             for(j = 1; j < obstacleGrid[0].size(); j++){
30                 if(obstacleGrid[i][j] == 0)
31                     dp[i][j] = dp[i-1][j] + dp[i][j-1];
32                 else
33                     dp[i][j] = 0;
34             }
35         }
36         
37         return dp[i-1][j-1];
38         
39     }
40 };

可以进行代码优化，节省空间。。。以后再说吧

 

posted @ 2016-08-03 16:00  琴影  阅读(180)  评论(0)  编辑  收藏  举报