问题描述:
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.
原问题链接:https://leetcode.com/problems/unique-paths-ii/
问题分析
这个问题和前面的问题很类似,唯一一个差别就是因为有了矩阵obstacleGrid。它里面所有标记为1的元素在对应的表示路径数量的矩阵matrix里对应的值为0。而且,在最开始设定条件的时候,最旁边的一行和一列,也就是matrix[m - 1][i], matrix[i][n - 1],它们在从最底角向上遍历的时候,如果一碰到对应的obstacleGrid元素为1,则可以直接跳出循环。因为有了这么一个点,它前面的路径都没法通过,也就是路径数为0。
在实现里再结合前面的条件matrix[i][j] = matrix[i + 1][j] + matrix[i][j + 1],我们可以很容易实现如下的代码:
public class Solution { public int uniquePathsWithObstacles(int[][] obstacleGrid) { int m = obstacleGrid.length; int n = obstacleGrid[0].length; int[][] matrix = new int[m][n]; if(obstacleGrid[m - 1][n - 1] == 1) return 0; for(int i = n - 1; i >= 0; i--) { if(obstacleGrid[m - 1][i] == 1) break; matrix[m - 1][i] = 1; } for(int i = m - 1; i >= 0; i--) { if(obstacleGrid[i][n - 1] == 1) break; matrix[i][n - 1] = 1; } for(int i = m - 2; i >= 0; i--) { for(int j = n - 2; j >= 0; j--) { if(obstacleGrid[i][j] == 1) matrix[i][j] = 0; else matrix[i][j] = matrix[i + 1][j] + matrix[i][j + 1]; } } return matrix[0][0]; } }