【LeetCode】128. Longest Consecutive Sequence 解題報告

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128. Longest Consecutive Sequence / Medium

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 <= 10^5
  • -10^9 <= nums[i] <= 10^9

Solution

思路

先把所有的元素都放進 unordered_set (Hash Set)，每一個元素進來先檢查是不是存在，存在的話把他從 set 移除，避免之後重複判斷。
接著把數字往前、往後增加，檢查 hset 是否有這些數字，最後計算出長度，和 max 相比，更新 max 的數值。

效能

Complexity

  • Time Complexity: O(N)
  • Space Complexity: O(N)

LeetCode Result

Code

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class Solution {
public:
    int longestConsecutive(vector<int>& nums) {
        if(nums.empty()) return 0;
        unordered_set<int> hset(nums.begin(), nums.end());
        
        int maxLCS = INT_MIN;
        for(const auto& num: nums) {
            if(hset.count(num) != 0) {
                hset.erase(num);
                int next = num+1, prev = num-1;
                while(hset.count(next) != 0) hset.erase(next++);
                while(hset.count(prev) != 0) hset.erase(prev--);
                maxLCS = max(next-prev-1, maxLCS);
            }
        }
        return maxLCS;
    }
};