[Swift]LeetCode659. 分割数组为连续子序列 | Split Array into Consecutive 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：https://www.cnblogs.com/strengthen/p/10488832.html 
➤如果链接不是山青咏芝的博客园地址，则可能是爬取作者的文章。
➤原文已修改更新！强烈建议点击原文地址阅读！支持作者！支持原创！
★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★

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

class Solution {
    func isPossible(_ nums: [Int]) -> Bool {
        var pre = 0
        var preCount = 0
        var starts = [Int]()
        var anchor = 0
        for i in 0..<nums.count {
            let t = nums[i]
            if i == nums.count - 1 || nums[i+1] != t {
                let count = i - anchor + 1
                if pre != 0 && (t - pre) != 1 {
                    while preCount > 0 {
                        if pre < (2 + starts.removeFirst()) {
                            return false
                        }
                        preCount -= 1
                    }
                    pre = 0
                }

                if pre == 0 || (t - pre) == 1{
                    while preCount > count {
                        preCount -= 1
                        if (t-1) < (2 + starts.removeFirst()) {
                            return false
                        }
                    }

                    while preCount < count {
                        starts.append(t)
                        preCount += 1
                    }
                }

                pre = t
                preCount = count
                anchor = i+1
            }
        }

        while preCount > 0 {
            if nums[nums.count - 1] < (2 + starts.removeFirst()) {
                return false
            }
            preCount -= 1
        }
        return true
    }
}

---

Runtime: 712 ms

Memory Usage: 19.4 MB

class Solution {
    func isPossible(_ nums: [Int]) -> Bool {
        var freq:[Int:Int] = [Int:Int]()
        var need:[Int:Int] = [Int:Int]()
        for num in nums
        {
            freq[num,default:0] += 1
        }
        for num in nums
        {
            if freq[num,default:0] == 0
            {
                continue
            }
            else if need[num,default:0] > 0
            {
                need[num,default:0] -= 1
                need[num + 1,default:0] += 1
            }
            else if freq[num + 1,default:0] > 0 && freq[num + 2,default:0] > 0
            {
                freq[num + 1,default:0] -= 1
                freq[num + 2,default:0] -= 1
                need[num + 3,default:0] += 1
            }
            else
            {
                return false
            }
            freq[num,default:0] -= 1
        }
        return true
    }
}