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Given an unsorted array of integers, find the length of the longest consecutive elements sequence.
For example,
Given [100, 4, 200, 1, 3, 2],
The longest consecutive elements sequence is [1, 2, 3, 4]. Return its length: 4.
Your algorithm should run in O(n) complexity.
以题目给出的数组为例：对于100，先向下查找99没找到，然后向上查找101也没找到，那么连续长度是1，从哈希表中删除100；然后是4，向下查找找到3,2,1，向上没有找到5，那么连续长度是4，从哈希表中删除4,3,2,1。这样对哈希表中已存在的某个元素向上和向下查找，直到哈希表为空。算法相当于遍历了一遍数组，然后再遍历了一遍哈希表，复杂的为O(n)
1 class Solution { 2 public: 3 int longestConsecutive(vector<int> &num) { 4 5 unordered_set<int> hash; 6 unordered_set<int>::iterator it; 7 8 for(int i=0;i<num.size();i++) hash.insert(num[i]); 9 10 int count; 11 int result=0; 12 while(!hash.empty()) 13 { 14 count=1; 15 16 it=hash.begin(); 17 int num0=*it; 18 hash.erase(num0); 19 20 int num=num0+1; 21 while(hash.find(num)!=hash.end()) 22 { 23 count++; 24 hash.erase(num); 25 num++; 26 } 27 28 num=num0-1; 29 while(hash.find(num)!=hash.end()) 30 { 31 count++; 32 hash.erase(num); 33 num--; 34 } 35 36 if(result<count) result=count; 37 } 38 39 return result; 40 } 41 };
【leetcode】Longest Consecutive Sequence
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原文地址：http://www.cnblogs.com/reachteam/p/4194385.html
