-
Notifications
You must be signed in to change notification settings - Fork 27
/
Copy path078. 子集.js
70 lines (61 loc) · 1.61 KB
/
078. 子集.js
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
/**
* Created by admin on 2018/8/3.
*/
/**给定一组不含重复元素的整数数组 nums,返回该数组所有可能的子集(幂集)。
* 说明:解集不能包含重复的子集。
*
* 示例:
* 输入: nums = [1,2,3]
* 输出:
* [
* [3],
* [1],
* [2],
* [1,2,3],
* [1,3],
* [2,3],
* [1,2],
* []
* ]
*/
/**很显然是077的变种,思路同077
* @param {number[]} nums
* @return {number[][]}
*/
var subsets = function(nums) {
nums.sort((a, b) => a - b);
var allGroup = [];
allGroup.push([]);
function combine(nums, k) {
var temp = [];
for (var i = 0; i < k; i++) {
temp.push(nums[i]);
}
allGroup.push(temp.concat([]));
var last = nums[nums.length - 1];
while (true) {
// 对尾数递增
while (temp[k - 1] < last) {
temp[k - 1] = nums[nums.indexOf(temp[k - 1]) + 1];
allGroup.push(temp.slice(0));
}
for (var i = k - 1; i >= 0; i--) {
var num = temp[i];
if (num < nums[nums.length - (k - i)]) {
var index = nums.indexOf(temp[i]);
for (var j = i;j < k; j++) {
temp[j] = nums[++index];
}
allGroup.push(temp.slice(0));
break;
}
}
if (i <= 0 && temp[0] === nums[nums.length - k]) return allGroup;
}
}
for (var i = 0; i < nums.length; i++) {
combine(nums, i + 1);
}
return allGroup;
};
console.log(subsets([0, 1, 4]))