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Unique Paths II

时间:2014-03-12 16:10:39      阅读:471      评论:0      收藏:0      [点我收藏+]

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.

简单动态规划,注意特殊情况,即第一个元素为1.

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class Solution {
public:
    int uniquePathsWithObstacles(vector<vector<int> > &obstacleGrid) {
        vector<int> p(obstacleGrid.size(),0);
        if(obstacleGrid[0][0] == 1)return 0;
        p[0] = 1;
        for(int i = 1 ; i <obstacleGrid.size();i++)
        if(obstacleGrid[i][0] == 0&&p[i-1] == 1) p[i] = 1;
        else p[i] = 0 ;
 
        for(int i = 1 ; i <obstacleGrid[0].size();i++)
        {
            if(obstacleGrid[0][i] == 1) p[0] = 0;
            for(int j = 1 ; j < obstacleGrid.size();j++)
            {
                if(obstacleGrid[j][i] == 1)p[j] = 0;
                else p[j] += p[j-1];
            }
        }
        return p[obstacleGrid.size()-1];
    }
};

  

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Unique Paths II

原文:http://www.cnblogs.com/pengyu2003/p/3595774.html

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