63. Unique Paths II
2017-03-06 17:00
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题意: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.
The total number of unique paths is 2.
Note: m and n will be at most 100.
思路:这题和之前的62. Unique Paths差不多,就多了个障碍点。具体思路还是一样,但是这里需要把障碍点的结果值置为0,所以需要做一些修改:
我的方法利用几个if判断来对第一行和第一列的结果进行一个选择,如果之前有障碍,则第一行/列之后的都为0,当然,后来看到更简单的方法,直接多开一行和一列的空间,全赋0,和62题的做法就基本一致了:
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.
思路:这题和之前的62. Unique Paths差不多,就多了个障碍点。具体思路还是一样,但是这里需要把障碍点的结果值置为0,所以需要做一些修改:
class Solution { public: int uniquePathsWithObstacles(vector<vector<int>>& obstacleGrid) { int m = obstacleGrid.size(); int n = obstacleGrid[0].size(); int result[m] ={1}; for (int i=0;i<m;i++){ for (int j = 0;j<n;j++){ if (obstacleGrid[i][j]==1) result[i][j]=0; else if (i==0 and j!=0) result[i][j]=result[i][j-1]; else if (i!=0 and j==0) result[i][j]=result[i-1][j]; else if (i>0 and j>0) result[i][j]=result[i-1][j]+result[i][j-1]; } } return result[m-1][n-1]; } };
我的方法利用几个if判断来对第一行和第一列的结果进行一个选择,如果之前有障碍,则第一行/列之后的都为0,当然,后来看到更简单的方法,直接多开一行和一列的空间,全赋0,和62题的做法就基本一致了:
class Solution { public: int uniquePathsWithObstacles(vector<vector<int>>& obstacleGrid) { int m = obstacleGrid.size(), n = obstacleGrid[0].size(); vector<vector<int> > dp(m + 1, vector<int> (n + 1, 0)); dp[0][1] = 1; for (int i = 1; i <= m; i++) for (int j = 1; j <= n; j++) if (!obstacleGrid[i - 1][j - 1]) dp[i][j] = dp[i - 1][j] + dp[i][j - 1]; return dp[m] ; } };
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