-
Notifications
You must be signed in to change notification settings - Fork 0
Expand file tree
/
Copy pathuniquePathsII.js
More file actions
81 lines (70 loc) · 2.24 KB
/
Copy pathuniquePathsII.js
File metadata and controls
81 lines (70 loc) · 2.24 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
74
75
76
77
78
79
80
81
// 63. Unique Paths II
// Medium
// 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
/**
* @param obstacleGrid: A list of lists of integers
* @return: An integer
*/
const uniquePathsWithObstacles = obstacleGrid => {
const n = obstacleGrid.length
if (!n) return 0
const m = obstacleGrid[0].length
if (!m) return 0
// 创建初始元素都为 0 的 m * n 的辅助二维数组
const dpArr = [], insideArr = []
for (let i = 0; i < m; i++) {
insideArr[i] = 0
}
for (let i = 0; i < n; i++) {
dpArr[i] = insideArr.slice() // 注意！！ 每次都用一个新的副本！
}
// 初始化
for (let i = 0; i < n; i++) {
if (obstacleGrid[i][0] != 1) {
dpArr[i][0] = 1;
} else {
break; // 避开空格
}
}
for (let i = 0; i < m; i++) {
if (obstacleGrid[0][i] != 1) {
dpArr[0][i] = 1;
} else {
break;
}
}
// 计算各有多种可能性的点
for (let i = 1; i < n; i++) {
for (let j = 1; j < m; j++) {
if (obstacleGrid[i][j] === 1) {
dpArr[i][j] = 0
} else {
dpArr[i][j] = dpArr[i - 1][j] + dpArr[i][j - 1]
}
}
}
return dpArr[n - 1][m - 1]
}
/**
* tip:
* 创建初始元素都为 0 的 m * n 的辅助二维数组时，一定注意，内层数组每次都要用新的副本！
* 否则，都成了一个相同的引用。外层数组里小数组（指针存储的地址都指向）为同一个实际数组对象。
* 带来的问题是，改其中一个小数组，其他小数组也都跟着改了。
*/