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.
Arraythinking：
（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 maxlen
int 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;
}
};leetcode || 128、Longest Consecutive Sequence
原文地址：http://blog.csdn.net/hustyangju/article/details/45331019
