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:

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:

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

 

Constraints:

• 0 <= nums.length <= 105
• -109 <= nums[i] <= 109

Solutions

Solution 1: Sorting

First, we sort the array, then use a variable $t$ to record the current length of the consecutive sequence, and a variable $ans$ to record the length of the longest consecutive sequence.

Next, we start traversing the array from index $i=1$. For the current element $nums[i]$:

• If $nums[i] = nums[i-1]$, it means the current element is repeated and does not need to be considered.
• If $nums[i] = nums[i-1] + 1$, it means the current element can be appended to the previous consecutive sequence to form a longer consecutive sequence. We update $t = t + 1$, and then update the answer $ans = \max(ans, t)$.
• Otherwise, it means the current element cannot be appended to the previous consecutive sequence, and we reset $t$ to $1$.

Finally, we return the answer $ans$.

The time complexity is $O(n \times \log n)$, and the space complexity is $O(\log n)$. Here, $n$ is the length of the array.

Python3JavaC++GoTypeScriptRustJavaScript

class Solution:
    def longestConsecutive(self, nums: List[int]) -> int:
        n = len(nums)
        if n < 2:
            return n
        nums.sort()
        ans = t = 1
        for a, b in pairwise(nums):
            if a == b:
                continue
            if a + 1 == b:
                t += 1
                ans = max(ans, t)
            else:
                t = 1
        return ans

class Solution {
    public int longestConsecutive(int[] nums) {
        int n = nums.length;
        if (n < 2) {
            return n;
        }
        Arrays.sort(nums);
        int ans = 1, t = 1;
        for (int i = 1; i < n; ++i) {
            if (nums[i] == nums[i - 1]) {
                continue;
            }
            if (nums[i] == nums[i - 1] + 1) {
                ans = Math.max(ans, ++t);
            } else {
                t = 1;
            }
        }
        return ans;
    }
}

class Solution {
public:
    int longestConsecutive(vector<int>& nums) {
        int n = nums.size();
        if (n < 2) {
            return n;
        }
        sort(nums.begin(), nums.end());
        int ans = 1, t = 1;
        for (int i = 1; i < n; ++i) {
            if (nums[i] == nums[i - 1]) {
                continue;
            }
            if (nums[i] == nums[i - 1] + 1) {
                ans = max(ans, ++t);
            } else {
                t = 1;
            }
        }
        return ans;
    }
};

func longestConsecutive(nums []int) int {
    n := len(nums)
    if n < 2 {
        return n
    }
    sort.Ints(nums)
    ans, t := 1, 1
    for i, x := range nums[1:] {
        if x == nums[i] {
            continue
        }
        if x == nums[i]+1 {
            t++
            ans = max(ans, t)
        } else {
            t = 1
        }
    }
    return ans
}

function longestConsecutive(</