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

★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★
➤微信公众号:山青咏芝(shanqingyongzhi)
➤博客园地址:山青咏芝(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 
➤若是连接不是山青咏芝的博客园地址,则多是爬取做者的文章。
➤原文已修改更新!强烈建议点击原文地址阅读!支持做者!支持原创!
★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★html

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 .git

Example:github

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:数组

  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.

给定一个整型数组, 你的任务是找到全部该数组的递增子序列,递增子序列的长度至少是2。微信

示例:app

输入: [4, 6, 7, 7]
输出: [[4, 6], [4, 7], [4, 6, 7], [4, 6, 7, 7], [6, 7], [6, 7, 7], [7,7], [4,7,7]]

说明:ide

  1. 给定数组的长度不会超过15。
  2. 数组中的整数范围是 [-100,100]。
  3. 给定数组中可能包含重复数字,相等的数字应该被视为递增的一种状况。

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 }

 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 }
相关文章
相关标签/搜索