leetcode128 Longest Consecutive Sequence

"""
Given an unsorted array of integers, find the length of the longest consecutive elements sequence.
Your algorithm should run in O(n) complexity.
Example:
Input: [100, 4, 200, 1, 3, 2]
Output: 4
Explanation: The longest consecutive elements sequence is [1, 2, 3, 4]. Therefore its length is 4.
"""
"""
解法一：关键用了自查找的想法，找到以后删除，节省时间
将nums用set处理一下，然后看成一个队列
首先将nums[0]出队列
令 j = nums[0]+1 ,判断nums[0]右边的连续数字是否在nums中,
如果存在：将nums[j]从数组中移除remove， j=j+1, 数字增大继续检查是否在nums中，
令 i = nums[0]-1, 判断nums[0]左边的连续数字是否在nums中,
如果存在：将nums[i]从数组中移除remove， i=i-1, 数字减小继续检查是否在nums中
用res=max(res, j-i-1)记录最大长度
"""
class Solution1:
    def longestConsecutive(self, nums):
        nums = set(nums)
        res = 0
        while nums:
            n = nums.pop()
            j = n+1 #！！！这种自查找式的思路很新颖
            while j in nums:
                nums.remove(j)
                j += 1
            i = n-1 #！！！
            while i in nums:
                nums.remove(i)
                i -= 1
            res = max(res, j-i-1)
        return res
"""
利用dict()，以数组中每个值为key，以该值往后连续序列为value，默认为1
并将i+1遍历后的数值删去
"""
class Solution2:
    def longestConsecutive(self, nums):
        d = dict()
        res = 0
        for num in nums:
            d[num] = 1
        for num in nums:
            if num in d:
                temp = num+1
                while temp in d:
                    d[num] += d[temp]
                    d.pop(temp)
                    temp += 1
                res = max(res, d[num])
        return res