LeetCode 128. Longest Consecutive Sequence

Description

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.

my program

思路：如果允许O(nlogn)的复杂度，那么可以先排序，可是本题要求O(n)。
由于序列里的元素是无序的，又要求O(n)，首先要想到用哈希表。

方法一

先排序，然后遍历找出最长递归序列；时间复杂度为O(nlogn)

class Solution {
public:
    int longestConsecutive(vector<int>& nums) {
        if(nums.empty()) return 0;
        sort(nums.begin(), nums.end());
        int max = 1;
        int count = 1;
        for(int i = 1; i<nums.size(); i++)
        {
            if(nums[i] == nums[i-1])
                continue;
            if((nums[i]-1) == nums[i-1])
                count++;
            else
                count = 1;
            if(max < count)
                max = count;
        }
        return max;
    }
};

Submission Details
68 / 68 test cases passed.
Status: Accepted
Runtime: 13 ms

方法二

用哈希表， 时间复杂度为O(n)

class Solution {
public:
    int longestConsecutive(vector<int>& nums) {
        if(nums.empty()) return 0;
        map<int,int> m1;
        int max = 1;
        int i = 0;
        for(; i<nums.size(); i++)
        {
            m1[nums[i]] = i;
        }
        while(!m1.empty())
        {
            int count = 1;
            i = m1.begin()->first;
            while(m1.find(i+1) != m1.end())
            {
                m1.erase(i+1);
                count++;
                i++;
            }

            i = m1.begin()->first;
            while(m1.find(i-1) != m1.end())
            {
                m1.erase(i-1);
                count++;
                i--;
            }
            m1.erase(m1.begin());
            if(max < count)
                max = count;
        }
        return max;
    }
};

Submission Details
68 / 68 test cases passed.
Status: Accepted
Runtime: 9 ms