[Leetcode]Longest Consecutive Sequence

dyllanzhou

于 2015-12-17 20:06:59 发布

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分类专栏： Union Find Array 文章标签： leetcode 算法

本文链接：https://blog.youkuaiyun.com/dyllanzhou/article/details/50346117

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本文介绍了一种寻找整数数组中最长连续元素序列的算法。通过两种方法实现：一种是排序后遍历数组，时间复杂度为O(n log n)，空间复杂度为O(1)；另一种是使用哈希集合进行扫描，时间复杂度为O(n)，空间复杂度为O(n)。这两种方法均可高效地解决此问题。

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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.

class Solution {
public:
    /*algorithm: native way, 
        sort and counting
        time O(nlgn) space O(1)
    */
    int longestConsecutive(vector<int>& nums) {
        int n = nums.size();
        if(n < 2)return n;
        int cnt = 1,maxCnt = 1;
        sort(nums.begin(),nums.end());
        for(int i = 1;i < n;i++){
            if(nums[i-1]+1 ==nums[i])cnt++;
            else if(nums[i-1] == nums[i])continue;
            else cnt = 1;
            maxCnt = max(cnt,maxCnt);
        }
        return maxCnt;
    }
};

class Solution {
public:
    /*algorithm: 
        scan and hash
        time O(n) space O(n)
    */
    int longestConsecutive(vector<int>& nums) {
        unordered_set<int>tbl;
        for(int i = 0;i < nums.size();i++){
            tbl.insert(nums[i]);
        }
        int maxCnt = 0,cnt = 0;
        for(auto it = tbl.begin(); it != tbl.end();it++){
            int cur = *it;
            int prev = cur-1,next=cur+1;
            cnt = 1;
            while(tbl.count(prev)){
                 tbl.erase(prev);
                 prev--;
                 cnt++;
            }
            while(tbl.count(next)){
                tbl.erase(next);
                next++;
                cnt++;
            }
            maxCnt = max(maxCnt,cnt);
        }
        return maxCnt;
    }
};