LeetCode 128: Longest Consecutive Sequence

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博客围绕LeetCode 128题Longest Consecutive Sequence展开，该题要求返回未排序整数数组中最长连续元素序列的长度。介绍了解法1，利用unordered_set，通过关键判断保证时间复杂度为O(n)；还提及解法3，用线性建堆法建最小堆统计最大连续个数，时间复杂度为O(nlogn)，解法2待确定。

128. Longest Consecutive Sequence

Hard

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

Follow up: Could you implement the O(n) solution?

Example 1:

Input: nums = [100,4,200,1,3,2]
Output: 4
Explanation: The longest consecutive elements sequence is [1, 2, 3, 4]. Therefore its length is 4.

Example 2:

Input: nums = [0,3,7,2,5,8,4,6,0,1]
Output: 9

Constraints:

0 <= nums.length <= 104
-109 <= nums[i] <= 109

解法1：unordered_set
参考的花生酱的网页
https://zxi.mytechroad.com/blog/hashtable/leetcode-128-longest-consecutive-sequence/
这里最关键的是 if (us.find(nums[i] - 1) == us.end())
这样当输入是[100,4,200,1,3,2]，我们碰到4，3，2就不会再进if了，这样保证时间复杂度O(n)。

class Solution {
public:
    int longestConsecutive(vector<int>& nums) {
        unordered_set us(nums.begin(), nums.end());
        int n = nums.size();
        int res = 0;
        for (int i = 0; i < n; ++i) {
            if (us.find(nums[i] - 1) == us.end()) {
                int count = 0;
                int candidate = nums[i];
                while(us.find(candidate) != us.end()) {
                    count++;
                    candidate++;
                }
                res = max(res, count);
            }
        }

        return res;
    }
};

解法2：TBD

解法3：我有个想法是用线性建堆法把nums[]建成一个最小堆，然后逐个pop，统计其中最大连续的个数。不过这样也是O(nlogn)。