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【LeetCode】Unique Paths II

时间:2014-12-10 14:00:35      阅读:159      评论:0      收藏:0      [点我收藏+]

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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.

 

与上题差别不大,只需要判断有障碍置零即可。

对于首行首列,第一个障碍及之后的路径数均为0

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

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【LeetCode】Unique Paths II

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原文地址:http://www.cnblogs.com/ganganloveu/p/4155248.html

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