[Swift]LeetCode491. 遞增子序列 | Increasing Subsequences

★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★
➤微信公衆號：山青詠芝（shanqingyongzhi）
➤博客園地址：山青詠芝（https://www.cnblogs.com/strengthen/）
➤GitHub地址：https://github.com/strengthen/LeetCode
➤原文地址：http://www.javashuo.com/article/p-bkjrrnfg-me.html 
➤若是連接不是山青詠芝的博客園地址，則多是爬取做者的文章。
➤原文已修改更新！強烈建議點擊原文地址閱讀！支持做者！支持原創！
★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★

Given an integer array, your task is to find all the different possible increasing subsequences of the given array, and the length of an increasing subsequence should be at least 2 .

Example:

Input: [4, 6, 7, 7]
Output: [[4, 6], [4, 7], [4, 6, 7], [4, 6, 7, 7], [6, 7], [6, 7, 7], [7,7], [4,7,7]] 

Note:

1. The length of the given array will not exceed 15.
2. The range of integer in the given array is [-100,100].
3. The given array may contain duplicates, and two equal integers should also be considered as a special case of increasing sequence.

---

給定一個整型數組, 你的任務是找到全部該數組的遞增子序列，遞增子序列的長度至少是2。

示例:

輸入: [4, 6, 7, 7]
輸出: [[4, 6], [4, 7], [4, 6, 7], [4, 6, 7, 7], [6, 7], [6, 7, 7], [7,7], [4,7,7]]

說明:

1. 給定數組的長度不會超過15。
2. 數組中的整數範圍是 [-100,100]。
3. 給定數組中可能包含重複數字，相等的數字應該被視爲遞增的一種狀況。

---

Runtime: 472 ms

Memory Usage: 17.3 MB

 1 class Solution {
 2     func findSubsequences(_ nums: [Int]) -> [[Int]] {
 3         var res:Set<[Int]> = Set<[Int]>()
 4         var cur:[[Int]] = [[]]
 5         for i in 0..<nums.count
 6         {
 7             var n:Int = cur.count
 8             for j in 0..<n
 9             {
10                 if !cur[j].isEmpty && cur[j].last! > nums[i]
11                 {
12                     continue
13                 }
14                 cur.append(cur[j])
15                 cur[cur.count - 1].append(nums[i])
16                 if cur[cur.count - 1].count >= 2
17                 {
18                     res.insert(cur.last!)
19                 }
20             }
21         }
22         return [[Int]](res)
23     }
24 }

---

15810 kb

 1 class Solution {
 2     func findSubsequences(_ nums: [Int]) -> [[Int]] {
 3         guard nums.count > 0 else { return [[Int]]() }
 4         
 5         var visit = [Bool](repeating: false, count: nums.count)
 6         var res = [[Int]]()
 7         
 8         dfs(nums, 0, &res, &visit, getPrev(nums), [Int]())
 9         
10         return res
11     }
12     
13     private func dfs(_ nums: [Int], _ index: Int, _ res: inout [[Int]], _ visit: inout [Bool], _ prev: [Int], _ list: [Int]) {
14         if list.count > 1 {
15             res.append(list)
16         }
17         
18         var list = list
19         
20         for i in index..<nums.count {
21             if index >= 1, nums[i] < nums[index-1] {
22                 continue
23             }
24             
25             if prev[i] != -1, visit[prev[i]] == false, (list.count == 0 || prev[i] > index-1) {
26                 continue
27             } 
28             
29             list.append(nums[i])
30             visit[i] = true
31             dfs(nums, i+1, &res, &visit, prev, list)
32             visit[i] = false
33             list.removeLast()
34         }
35     }
36     
37     private func getPrev(_ nums: [Int]) -> [Int] {
38         var map = [Int: Int]()
39         var prev = [Int](repeating: -1, count: nums.count)
40         
41         for i in 0..<nums.count {
42             if let value = map[nums[i]] {
43                 prev[i] = value
44             } 
45             map[nums[i]] = i
46         }
47         
48         return prev
49     }
50 }