[LeetCode] 128. Longest Consecutive Sequence

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题：https://leetcode.com/problems/longest-consecutive-sequence/description/

题目

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.
#思路
两种方式
##1.
利用sort先对nums进行排序，然后遍历一次，如果当前遍历的和上次访问的数据相同则跳过，否则，若当前数据比上次访问的数据大一，则不同。

##2.
利用 set对nums进行set操作。

##Solution
https://leetcode.com/problems/longest-consecutive-sequence/solution/

#code
1.

class Solution:def longestConsecutive(self, nums):""":type nums: List[int]:rtype: int"""if len(nums) == 0:return 0nums.sort()longest_streak = 1current_streak = 1for i in range(1,len(nums)):if nums[i]!=nums[i-1]:if nums[i] == nums[i-1]+1:current_streak +=1else:longest_streak = max(longest_streak,current_streak)current_streak =1return max(longest_streak,current_streak)

class Solution:def longestConsecutive(self, nums):""":type nums: List[int]:rtype: int"""longest_streak = 0numsset = set(nums)for num in nums:if num-1 not in numsset:current_num = numcurrent_streak = 1while current_num + 1 in numsset:current_num +=1current_streak +=1longest_streak = max(longest_streak,current_streak)return longest_streak

第二版

1. 将 nums中元素放入 set中。
2. 遍历set中元素，若 该元素+1的数 在set中，那么不断遍历递增元素，求出最长序列；若 元素-1的数 在 set中，那么就不求最长序列。因为当遍历到 该元素-1 的数时，统计最长序列会包含以该元素为起点的最长序列。

class Solution {public int longestConsecutive(int[] nums) {Set<Integer> set = new HashSet<>();for(int num:nums)set.add(num);int max = 0;for(int num : nums){if(set.contains(num -1))continue;int start =num;while(set.contains(num+1)){num++;}max = Math.max(max,num - start +1);}return max;}
}

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