Unique Paths II

来源：Leetcode

原帖：http://oj.leetcode.com/problems/unique-paths-ii/

题目：
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.

代码：

 class Solution {  
 public:  
   int uniquePathsWithObstacles(vector<vector<int> > &obstacleGrid) {  
     int m = obstacleGrid.size(), n = obstacleGrid[0].size();  
     int dp[m][n];  
     if (obstacleGrid[0][0] == 1) return 0;  
     dp[0][0] = 1;  
     for (int i = 1; i < m; i++)  
       dp[i][0] = obstacleGrid[i][0] == 1 ? 0 : dp[i - 1][0];        
     for (int j = 1; j < n; j++)  
       dp[0][j] = obstacleGrid[0][j] == 1 ? 0 : dp[0][j - 1];        
     for (int i = 1; i < m; i++) {  
       for (int j = 1; j < n; j++) {  
         dp[i][j] = obstacleGrid[i][j] == 1 ? 0: dp[i - 1][j] + dp[i][j - 1];                
       }  
     }   
     return dp[m - 1][n - 1];  
   }  
 };  
   
 // space complexity O(n).   
 class Solution {  
 public:  
   int uniquePathsWithObstacles(vector<vector<int> > &obstacleGrid) {  
     if (obstacleGrid.size() == 0 || obstacleGrid[0].size() == 0) return 0;  
     int m = obstacleGrid.size(), n=obstacleGrid[0].size();  
     vector<int> dp(n + 1, 0);  
     dp[1] = obstacleGrid[0][0] == 1 ? 0 : 1;  
     for (int i = 0; i < m; i++) {  
       for (int j = 1; j <= n; j++){  
         dp[j] = obstacleGrid[i][j-1] == 1 ? 0 : dp[j] + dp[j-1];  
       }  
     }  
     return dp[n];  
   }  
 };