problem：
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.
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Array
题意：在乱序数组中寻找最长的连续子序列（没有顺序要求）
thinking：
（1）要求时间复杂度为O（n），只能使用hashtable了。
（2）unordered_map底层就是用hashtable实现的，可以拿来当做hashtable使用
code：
class Solution {public:int longestConsecutive(vector<int> &num) {unordered_map<int,int> hashtable;for(int i = 0;i < num.size();i++) {if(hashtable.find(num[i]) != hashtable.end())continue;int minus_1 = num[i] – 1;int plus_1 = num[i] + 1;unordered_map<int,int>::iterator minus_1iter,plus_1iter;minus_1iter = hashtable.find(minus_1);plus_1iter = hashtable.find(plus_1);if(minus_1iter != hashtable.end() && plus_1iter != hashtable.end()) {hashtable[num[i]] = hashtable[minus_1] + hashtable[plus_1] + 1;hashtable[num[i] – hashtable[minus_1]] = hashtable[num[i]];hashtable[num[i] + hashtable[plus_1]] = hashtable[num[i]];}else if(minus_1iter == hashtable.end() && plus_1iter == hashtable.end()) {hashtable[num[i]] = 1;}else if(minus_1iter != hashtable.end()) {hashtable[num[i]] = hashtable[minus_1] + 1;hashtable[num[i] – hashtable[minus_1]] = hashtable[num[i]];}else {hashtable[num[i]] = hashtable[plus_1] + 1;hashtable[num[i] + hashtable[plus_1]] = hashtable[num[i]];}}//find the maxlenint ans = INT_MIN;for(unordered_map<int,int>::iterator iter = hashtable.begin();iter != hashtable.end();++iter) {if(iter->second > ans)ans = iter->second;}return ans;}};
，未曾失败的人恐怕也未曾成功过。
