-
Notifications
You must be signed in to change notification settings - Fork 2
Expand file tree
/
Copy pathUniquePathsII.java
More file actions
73 lines (67 loc) · 2.04 KB
/
UniquePathsII.java
File metadata and controls
73 lines (67 loc) · 2.04 KB
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
package leetcode;
/**
* @author eko
* @date 2018/10/23 7:34 PM
*
* A robot is located at the top-left corner of a m x n grid (marked 'Start' in the diagram below).
*
* The robot can only move either down or right at any point in time. The robot is trying to reach the bottom-right corner of the grid (marked 'Finish' in the diagram below).
*
* 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.
*
* Note: m and n will be at most 100.
*
* Example 1:
*
* Input:
* [
* [0,0,0],
* [0,1,0],
* [0,0,0]
* ]
* Output: 2
* Explanation:
* There is one obstacle in the middle of the 3x3 grid above.
* There are two ways to reach the bottom-right corner:
* 1. Right -> Right -> Down -> Down
* 2. Down -> Down -> Right -> Right
*/
public class UniquePathsII {
public static void main(String[] args) {
int[][] obstacleGrid =
{
{0, 0, 0},
{0, 1, 0},
{0, 0, 0}
};
int res = new UniquePathsII().uniquePathsWithObstacles(obstacleGrid);
System.out.println(res);
}
public int uniquePathsWithObstacles(int[][] obstacleGrid) {
int[][] matrix = new int[obstacleGrid.length][obstacleGrid[0].length];
for (int i = 0; i < matrix.length; i++) {
if (obstacleGrid[i][0] == 1) {
break;
}
matrix[i][0] = 1;
}
for (int i = 0; i < matrix[0].length; i++) {
if (obstacleGrid[0][i] == 1) {
break;
}
matrix[0][i] = 1;
}
for (int i = 1; i < matrix.length; i++) {
for (int j = 1; j < matrix[0].length; j++) {
if (obstacleGrid[i][j] != 1) {
matrix[i][j] = matrix[i - 1][j] + matrix[i][j - 1];
}
}
}
return matrix[obstacleGrid.length - 1][obstacleGrid[0].length - 1];
}
}