【LeetCode】128. Longest Consecutive Sequence

128. Longest Consecutive Sequence

介绍

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

题意：给定一个无序数组，数组中可能存在重复元素，数组中最长的连续序列长度。

解答

方法一：

这个方法采用了map来帮助完成功能。在遍历过程中，map键值是元素的值，如果此时这个元素是一个序列中的边界，如有序列，1到3，map键值对应的值就是这个序列的长度，即map[1] = 3，且map[3] = 3。

class Solution {public:    int longestConsecutive(vector<int>& nums)     {        unordered_map<int,int> help;        int res = 0;        for(auto val:nums)        {            if(help[val])   continue;            int left = help[val-1];            int right = help[val+1];            res = max(res,help[val-left] = help[val+right] = help[val] = left+right+1);        }        return res;    }};

方法二

并查集。这个方法时间复杂度和空间复杂度都很高，并不推荐，写在这里只是为了让自己学习并查集这个算法。

class Union_Set{public:    Union_Set(int n):mark(n)    {        for(int i = 0; i < n; ++i)  mark[i] = i;    }    void make_union(int a,int b)    {        int aroot = find(a);        int broot = find(b);        mark[aroot] = broot;    }    int find (int a)    {        while(a != mark[a])            a = mark[a];        return a;    }    int max_union()    {        vector<int> temp(mark.size());        int res = 0;        for(int i = 0; i < mark.size(); ++i)        {            res = max( res,++temp[find(i)] );        }        for(auto val:mark)            cout << val << " ";        cout << endl;        return res;    }private:    vector<int> mark;};class Solution {public:    int longestConsecutive(vector<int>& nums)     {        //并查集方法        unordered_map<int,int> help;        Union_Set record(nums.size());        for(int i = 0; i < nums.size(); ++i)        {            if(help.find(nums[i]) != help.end())    continue;            help[nums[i]] = i;            if(help.find(nums[i]-1) != help.end())  record.make_union(help[nums[i]-1],i);            if(help.find(nums[i]+1) != help.end())  record.make_union(help[nums[i]+1],i);        }        return record.max_union();    }};