[leetcode] Longest Consecutive Sequence

最新推荐文章于 2022-03-06 23:29:26 发布

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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，并通过使用unordered_map实现了O(n)的时间复杂度。该算法适用于未排序的整数数组，通过查找每个元素的左右邻居来确定连续序列的长度。

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.

这道题目要求了o(n)的时间复杂度，其实没有什么特别巧妙的做法，也就是对set容器的使用

int longestConsecutive(vector<int> &num) {
    unordered_map<int,int> s;
    int maxlen = 0;
    for(vector<int>::iterator it = num.begin(); it != num.end(); it++) s[*it] = 1;
    //each element in the array can be accessed at most twice,
    //therefore the amortized running time is still o(n).
    for(vector<int>::iterator it = num.begin(); it != num.end(); it++){
        if(s[*it] == 1){
            s[*it] = 0;
            int len = 1;
            int left = *it - 1;
            int right = *it + 1;

            while(s.find(left) != s.end() && s[left]){
                s[left--] = 0;
                len++;
            }
            while(s.find(right) != s.end() && s[right]){
                s[right++] = 0;
                len++;
            }
            if(maxlen < len) maxlen = len;
        }
    }
    return maxlen;
}