织梦手机端网站怎么做磁力搜索器

1.题目描述

2.思路

暴力解法是会超出时间限制的。首先将nums写成一个集合（哈希表），无限循环，对于集合顶端的元素向左and向右扩展最大长度，每访问一个元素就把它从集合里删掉，循环的结束条件是集合为空。

3.代码（Python3）

class Solution:def longestConsecutive(self, nums: List[int]) -> int:def neighbor_exist(num):left_num, right_num = num - 1, num + 1current_consecutive = 1while 1:if left_num in nums_set:nums_set.remove(left_num)current_consecutive += 1left_num -= 1else:breakwhile 1:if right_num in nums_set:nums_set.remove(right_num)current_consecutive += 1right_num += 1else:breakreturn current_consecutivenums_set = set(nums)longest_consecutive = 0while 1:if len(nums_set) == 0:return longest_consecutivecurrent_consecutive = neighbor_exist(nums_set.pop())longest_consecutive = max(longest_consecutive, current_consecutive)

4.执行情况

5.感想

看了官方题解发现了更简便的代码，直接遍历集合，如果当前元素不是当前序列的开始元素就continue，如果是就扩展长度直到扩展不了，如下：

class Solution:def longestConsecutive(self, nums: List[int]) -> int:nums_set = set(nums)longest_consecutive = 0for num in nums_set:if (num - 1) not in nums_set:current_consecutive = 1current_num = numwhile (current_num + 1) in nums_set:current_consecutive += 1current_num += 1longest_consecutive = max(longest_consecutive, current_consecutive)return longest_consecutive

但是这个代码执行出来的效果竟然不如我上面写的那个。不过也确实，官方题解是把全部元素都判断了一遍是不是开头，我是用完就删，确实应该相对高效。