[LeetCode]Longest Consecutive Sequence

最新推荐文章于 2025-03-19 23:17:40 发布

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本文介绍了一种使用C++实现的寻找最长连续整数序列的算法。通过哈希表记录数组元素，算法能在O(n)时间内找到最长连续序列。文章详细展示了如何遍历输入数组并利用两个哈希表来标记已访问元素及存储数组值。

class Solution {
public:
	int longestConsecutive(vector<int> &num) {
		// Start typing your C/C++ solution below
		// DO NOT write int main() function
		//O(n) to find the longest consecutive sequence so we need to use O(1) hash table
		//unordered_set is an appropriate one
		unordered_set<int> hashTable;
		for(int i = 0; i < num.size(); ++i)
			hashTable.insert(num[i]);
		//enumerate every hash value
		unordered_set<int> visitedTable;//keep the visit status of hash value
		int maxLen = 1;
		for (int i = 0; i < num.size(); ++i)
		{
			if(visitedTable.find(num[i]) != visitedTable.end()) continue;
			visitedTable.insert(num[i]);
			int curLen = 1;
			int nextNum = num[i]-1;//search left
			while (hashTable.find(nextNum) != hashTable.end())
			{
				visitedTable.insert(nextNum);
				nextNum--;
				curLen++;
			}
			nextNum = num[i]+1;//search right
			while (hashTable.find(nextNum) != hashTable.end())
			{
				visitedTable.insert(nextNum);
				nextNum++;
				curLen++;
			}
			//keep the maxLen updated
			maxLen = max(maxLen, curLen);
		}
		return maxLen;
	}
};

second time

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;
    }
};