LeetCode 128. Longest Consecutive Sequence 最长连续序列 (C++/Java)
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.
分析:
给定一个未排序的整数数组,找出最长连续序列的长度。
可以先对数组进行排序,然后遍历数组,判断数字是否连续来计算最大长度,不过由于排序,时间复杂度是O(nlogn),我们可以利用哈希表来存储数组元素,再遍历元素,当前元素为num时,如果num-1这个元素不在我们的集合中,就代表这个num可以作为序列的起始点,然后依次判断num++是否在集合中,更新当前序列最大长度,当出现num++不在集合中,也就是此时序列不再连续,更新全局最大长度,继续遍历数组,最后返回全局的最大长度即可。
程序:
C++
class Solution { public: int longestConsecutive(vector<int>& nums) { unordered_set<int> set(nums.begin(), nums.end()); int res = 0; for(int num:nums){ if(!set.count(num-1)){ int l = 1; while(set.count(++num)){ l++; } res = max(res, l); } } return res; } };
Java
class Solution { public int longestConsecutive(int[] nums) { if(nums.length == 0) return 0; int res = 0; Set<Integer> set = new HashSet<>(); for(int num:nums) set.add(num); for(int num:nums){ if(!set.contains(num-1)){ int l = 1; while(set.contains(++num)) l++; res = Math.max(l, res); } } return res; } }