128. Longest Consecutive Sequence

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128. Longest Consecutive Sequence

Description

Given an unsorted array of integers nums, return the length of the longest consecutive elements sequence.

You must write an algorithm that runs in O(n) time.

Example 1:

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Input: nums = [100,4,200,1,3,2]
Output: 4
Explanation: The longest consecutive elements sequence is `[1, 2, 3, 4]`. Therefore its length is 4.

Example 2:

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Input: nums = [0,3,7,2,5,8,4,6,0,1]
Output: 9

Constraints:

  • 0 <= nums.length <= 10^5
  • -10^9 <= nums[i] <= 10^9

Hints/Notes

Solution

Language: C++

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class Solution {
public:
    int longestConsecutive(vector<int>& nums) {
        int res = 0;
        unordered_set<int> s(nums.begin(), nums.end());
        for (int x : s) {
            if (s.contains(x - 1)) {
                continue;
            }
            int y = x + 1;
            while (s.contains(y)) {
                y++;
            }
            res = max(res, y - x);
        }
        return res;
    }
};

Sort solution(not correct, O(nlogn) time):

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class Solution {
public:
    int longestConsecutive(vector<int>& nums) {
        ranges::sort(nums);
        int res = 0, cur = 0, curVal = INT_MIN;
        for (int i = 0; i < nums.size(); i++) {
            if (nums[i] == curVal) {
                continue;
            }
            if (nums[i] == curVal + 1) {
                curVal++;
                cur++;
            } else {
                res = max(res, cur);
                curVal = nums[i];
                cur = 1;
            }
        }
        res = max(res, cur);
        return res;
    }
};