[LeetCode]Longest Consecutive Sequence

最新推荐文章于 2025-09-07 19:33:12 发布

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最新推荐文章于 2025-09-07 19:33:12 发布

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CC 4.0 BY-SA版权

文章标签： c/c++

原文链接：http://www.cnblogs.com/Rosanna/p/3597902.html

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。利用unordered_set实现O(1)查找，确保整体算法复杂度为O(n)。通过示例说明了如何找到[100,4,200,1,3,2]中长度为4的最长连续序列。

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.

思考：空间换时间，查找O(1)用unordered_set。

class Solution {
public:
    int longestConsecutive(vector<int> &num) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        unordered_set<int> existTable;
        for(int i = 0; i < num.size(); ++i) existTable.insert(num[i]);
        unordered_set<int> visitedTable;
        int mmax = INT_MIN;
        for(int i = 0; i < num.size(); ++i)
        {
            int curNum = num[i];
            if(visitedTable.find(curNum) != visitedTable.end()) continue;
            int cnt = 1;
            //search left
            int left = curNum;
            while(existTable.find(--left) != existTable.end()) cnt++, visitedTable.insert(left);
            int right = curNum;
            while(existTable.find(++right) != existTable.end()) cnt++, visitedTable.insert(right);
            mmax = max(cnt, mmax);
        }
        return mmax;
    }
};

转载于:https://www.cnblogs.com/Rosanna/p/3597902.html