LeetCode 128. 最长连续序列

排序+遍历（不符合时间复杂度要求）

class Solution {
public:
    int longestConsecutive(vector<int>& nums) {
        int len = nums.size();
        if(len==0)
            return 0;

        sort(nums.begin(),nums.end());

        for(auto it : nums)
            cout<<it<<" ";
        cout<<endl;

        int cur_len = 1;
        int res = 1;
        int cur_num = nums[0];
        for(int i=1; i<len; i++)
        {
            while(i<len && nums[i] == cur_num)
            {
                i++;
            }

            if(i<len && nums[i] == cur_num + 1)
            {
                cur_len += 1;
                res = max(res, cur_len);
            }
            else
            {
                cur_len = 1;
            }

            cur_num = nums[i];
        }

        return res;
    }
};

使用hash表

class Solution {
public:
    int longestConsecutive(vector<int>& nums) {
        unordered_set<int> myset(nums.begin(), nums.end());
        int res = 0;

        for(int num : myset)
        {
            // 如果num-1在集合中，说明已经被使用过，直接跳过
            if(myset.contains(num-1))
                continue;

            // 否则，获取以num为起点的序列长度
            int y = num + 1;
            while(myset.contains(y))
            {
                y++;
            }
            res = max(res, y-num);
        }
        return res;
    }
};