Unique Paths II <leetcode>

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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.

 

演算法：一開始理解錯了，以為到一個點可以從上下左右，寫了一個非常複雜的代碼，後來自己感覺不對，上網上一查才發現，原來還是到一個點只能從上或左，那這道題就非常簡單了，思路和第一個版本是一樣的，只是增加了障礙物，所以計算的時候，考慮一下障礙物就行了，代碼如下（動態規劃）：

 

 1 class Solution { 2 public: 3     int uniquePathsWithObstacles(vector<vector<int> > &obstacleGrid) { 4         vector<vector<int>>  f(obstacleGrid.size(),vector<int>(obstacleGrid[0].size())); 5         f[0][0]=obstacleGrid[0][0]==0?1:0; 6         for(int i=1;i<f.size();i++) 7         { 8             f[i][0]=obstacleGrid[i][0]==0?f[i-1][0]:0; 9         }10         for(int i=1;i<obstacleGrid[0].size();i++)11         {12             f[0][i]=obstacleGrid[0][i]==0?f[0][i-1]:0;13         }14         15         for(int i=1;i<obstacleGrid.size();i++)16         {17             for(int j=1;j<obstacleGrid[0].size();j++)18             {19                 f[i][j]=obstacleGrid[i][j]==0?(f[i-1][j]+f[i][j-1]):0;20             }21         }22         return f[f.size()-1][f[0].size()-1];23     }24 };

 

Unique Paths II <leetcode>