[Swift]LeetCode659. 分割数组为连续子序列 | Split Array into Consecutive Subsequences

You are given an integer array sorted in ascending order (may contain duplicates), you need to split them into several subsequences, where each subsequences consist of at least 3 consecutive integers. Return whether you can make such a split.

Example 1:

Input: [1,2,3,3,4,5]
Output: True
Explanation:
You can split them into two consecutive subsequences : 
1, 2, 3
3, 4, 5 

Example 2:

Input: [1,2,3,3,4,4,5,5]
Output: True
Explanation:
You can split them into two consecutive subsequences : 
1, 2, 3, 4, 5
3, 4, 5 

Example 3:

Input: [1,2,3,4,4,5]
Output: False 

Note:

1. The length of the input is in range of [1, 10000]

---

输入一个按升序排序的整数数组（可能包含重复数字），你需要将它们分割成几个子序列，其中每个子序列至少包含三个连续整数。返回你是否能做出这样的分割？ 

示例 1：

输入: [1,2,3,3,4,5]
输出: True
解释:
你可以分割出这样两个连续子序列 : 
1, 2, 3
3, 4, 5 

示例 2：

输入: [1,2,3,3,4,4,5,5]
输出: True
解释:
你可以分割出这样两个连续子序列 : 
1, 2, 3, 4, 5
3, 4, 5 

示例 3：

输入: [1,2,3,4,4,5]
输出: False 

提示：

1. 输入的数组长度范围为 [1, 10000]

---

620ms

 1 class Solution {
 2     func isPossible(_ nums: [Int]) -> Bool {
 3         var pre = 0
 4         var preCount = 0
 5         var starts = [Int]()
 6         var anchor = 0
 7         for i in 0..<nums.count {
 8             let t = nums[i]
 9             if i == nums.count - 1 || nums[i+1] != t {
10                 let count = i - anchor + 1
11                 if pre != 0 && (t - pre) != 1 {
12                     while preCount > 0 {
13                         if pre < (2 + starts.removeFirst()) {
14                             return false
15                         }
16                         preCount -= 1
17                     }
18                     pre = 0
19                 }
20 
21                 if pre == 0 || (t - pre) == 1{
22                     while preCount > count {
23                         preCount -= 1
24                         if (t-1) < (2 + starts.removeFirst()) {
25                             return false
26                         }
27                     }
28 
29                     while preCount < count {
30                         starts.append(t)
31                         preCount += 1
32                     }
33                 }
34 
35                 pre = t
36                 preCount = count
37                 anchor = i+1
38             }
39         }
40 
41         while preCount > 0 {
42             if nums[nums.count - 1] < (2 + starts.removeFirst()) {
43                 return false
44             }
45             preCount -= 1
46         }
47         return true
48     }
49 }

---

Runtime: 712 ms

Memory Usage: 19.4 MB

 1 class Solution {
 2     func isPossible(_ nums: [Int]) -> Bool {
 3         var freq:[Int:Int] = [Int:Int]()
 4         var need:[Int:Int] = [Int:Int]()
 5         for num in nums
 6         {
 7             freq[num,default:0] += 1
 8         }
 9         for num in nums
10         {
11             if freq[num,default:0] == 0
12             {
13                 continue
14             }
15             else if need[num,default:0] > 0
16             {
17                 need[num,default:0] -= 1
18                 need[num + 1,default:0] += 1
19             }
20             else if freq[num + 1,default:0] > 0 && freq[num + 2,default:0] > 0
21             {
22                 freq[num + 1,default:0] -= 1
23                 freq[num + 2,default:0] -= 1
24                 need[num + 3,default:0] += 1
25             }
26             else
27             {
28                 return false
29             }
30             freq[num,default:0] -= 1
31         }
32         return true
33     }
34 }