#124 Longest Consecutive Sequence

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本文介绍了一种使用哈希集合（set）实现的高效算法，该算法可在O(n)的时间复杂度内找到整数数组中最长的连续元素序列。通过实例演示了如何遍历集合并检查相邻元素的存在，以确定最长连续序列的长度。

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题目描述：

Given an unsorted array of integers, find the length of the longest consecutive elements sequence.

Have you met this question in a real interview?

Yes

Clarification

Your algorithm should run in O(n) complexity.

Example

Given [100, 4, 200, 1, 3, 2],
The longest consecutive elements sequence is [1, 2, 3, 4]. Return its length: 4.

题目思路：

这题可以用一个set来记录num中的每个element。然后遍历这个set，比如，set的某个number为tmp。那我们可以查看tmp + 1, tmp + 2, tmp + 3, ... etc.和tmp - 1, tmp - 2, tmp - 3, etc...是不是也在set中（同时在set中删去这些数），然后算出以tmp为中心的consecutive elements有多长。对于set中的每个数都这样操作，最后找出最长的那个答案。

Mycode（AC = 29ms）：

class Solution {
public:
    /**
     * @param nums: A list of integers
     * @return an integer
     */
    int longestConsecutive(vector<int> &num) {
        // write you code here
        unordered_set<int> helper;
        
        // put numbers into the set
        for (int i = 0; i < num.size(); i++) {
            if (helper.find(num[i]) == helper.end()) {
                helper.insert(num[i]);
            }
        }
        
        if (num.size() <= 1) return num.size();
        
        // check each number in the set, and check whether number++
        // and number-- are in the set, then get the local_count
        // for each number in the set
        int count = 1, local_count = 1;
        for (auto it = helper.begin(); it != helper.end(); it++) {
            int tmp = *it;
            local_count = 1;
            
            while (helper.find(tmp + 1) != helper.end()) {
                helper.erase(tmp + 1);
                tmp++;
                local_count++;
            }
            
            tmp = *it;
            while (helper.find(tmp - 1) != helper.end()) {
                helper.erase(tmp - 1);
                tmp--;
                local_count++;
            }
            
            count = max(count, local_count);
        }
        
        return count;
    }
};