63. Unique Paths II
2017-07-05 14:16
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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
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
Note: m and n will be at most 100.
题意:和Unique Paths 类似 只是多了个障碍的限制
思想:那么我们可以想办法把这题转化成 Unique Paths 来做
先判断第一列和第一行 如果位置为0 就把它标记为1 如果为1之后的位置全标记为0 因为都不可达
再来判断其它的位置,如果该位置是障碍为1时 就标记为0(能到达的路径为0)
如果为0时那么就按照Unique Paths的思想 标记为path[i-1][j]+path[i][j-1];
AC代码:时间O(n*m) 空间O(n*m)
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
1and
0respectively
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.
题意:和Unique Paths 类似 只是多了个障碍的限制
思想:那么我们可以想办法把这题转化成 Unique Paths 来做
先判断第一列和第一行 如果位置为0 就把它标记为1 如果为1之后的位置全标记为0 因为都不可达
再来判断其它的位置,如果该位置是障碍为1时 就标记为0(能到达的路径为0)
如果为0时那么就按照Unique Paths的思想 标记为path[i-1][j]+path[i][j-1];
AC代码:时间O(n*m) 空间O(n*m)
class Solution { public: int uniquePathsWithObstacles(vector<vector<int>>& obstacleGrid) { int m = obstacleGrid.size(); int n = obstacleGrid[0].size(); //1代表该位置能到达的一条路径 0表示0条路径 ///判断第一列 如果该位置为0 就标记1 如果该位置为1 那么该位置以及后面位置全为0 int flagx = 1; for(int i =0;i<m;i++) if(obstacleGrid[i][0]==0&&flagx) obstacleGrid[i][0] = 1; else{ obstacleGrid[i][0] = 0; flagx=0; } ///判断第一行 如果该位置为0 就标记1 如果该位置为1 那么该位置以及后面位置全为0 int flagy = 1; obstacleGrid[0][0] ^=1; //obstacleGrid[0][0] 起始位置 取原始值 再判断这一行每一位置的路径数 for(int i =0;i<n;i++) if(obstacleGrid[0][i]==0&&flagy) obstacleGrid[ 9bea 0][i] = 1; else{ obstacleGrid[0][i] = 0; flagy=0; } //和Unique Paths的思想一样 这里遍历的位置为1代表有障碍 不可达 所以设置可达路径为0 for(int i=1;i<m;i++){ for(int j=1;j<n;j++){ if(obstacleGrid[i][j]==0){ obstacleGrid[i][j] = obstacleGrid[i-1][j]+obstacleGrid[i][j-1]; }else{ obstacleGrid[i][j] = 0; } } } return obstacleGrid[m-1][n-1]; } };
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