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【leetcode刷题笔记】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.
题解:和http://www.cnblogs.com/sunshineatnoon/p/3798167.html非常相似,要注意两点:
- 有障碍的地方能走到的方法数目是0;
- 初始化结果矩阵的时候,如果第一行(列)的某个位置有障碍,那么这一行(列)该元素后面所有的元素都是0,不能置为1;比如给定障碍矩阵[1,0,0],那么初始化以后的矩阵应该为[0,0,0],而不是[0,1,1],因为(0,0)处有障碍,那么该矩阵的任何位置都是无法到达的。
代码如下:
1 public class Solution { 2 public int uniquePathsWithObstacles(int[][] obstacleGrid) { 3 int m = obstacleGrid.length; 4 int n = obstacleGrid[0].length; 5 if(m==0 && n == 0) 6 return 0; 7 8 int[][] PathNum = new int[m][n]; 9 for(int i = 0;i < m;i++)10 if(obstacleGrid[i][0] != 1)11 PathNum[i][0] = 1;12 else 13 break;14 for(int i = 0;i < n;i++)15 if(obstacleGrid[0][i] != 1)16 PathNum[0][i] = 1;17 else18 break;19 20 for(int i = 1;i < m;i++)21 for(int j = 1;j < n;j++){22 if(obstacleGrid[i][j] != 1)23 PathNum[i][j] = PathNum[i-1][j]+PathNum[i][j-1]; 24 }25 26 return PathNum[m-1][n-1];27 }28 }
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