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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.
1 class Solution { 2 public: 3 int uniquePathsWithObstacles(vector<vector<int>>& obstacleGrid) { 4 int m = obstacleGrid.size(); 5 int n = obstacleGrid[0].size(); 6 vector<vector<int>> dp(m, vector<int>(n, 1)); 7 for(int i = 0;i < m;++i) 8 { 9 for(int j = 0; j < n;++j) 10 { 11 if(obstacleGrid[i][j] == 1) 12 { 13 dp[i][j] = 0; 14 continue; 15 } 16 if(i == 0 && j == 0) dp[i][j] = 1; 17 else if(i == 0) dp[i][j] = dp[i][j-1]; 18 else if(j == 0) dp[i][j] = dp[i-1][j]; 19 else 20 dp[i][j] = dp[i-1][j] + dp[i][j-1]; 21 } 22 } 23 return dp[m-1][n-1]; 24 } 25 };
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原文地址：http://www.cnblogs.com/chdxiaoming/p/4786402.html
