leetcode 414. Third Maximum Number | set

Description

Given a non-empty array of integers, return the third maximum number in this array. If it does not exist, return the maximum number. The time complexity must be in O(n).

Example 1:
Input: [3, 2, 1]

Output: 1

Explanation: The third maximum is 1.
Example 2:
Input: [1, 2]

Output: 2

Explanation: The third maximum does not exist, so the maximum (2) is returned instead.
Example 3:
Input: [2, 2, 3, 1]

Output: 1

Explanation: Note that the third maximum here means the third maximum distinct number.
Both numbers with value 2 are both considered as second maximum.

Discuss

因为有去重的要求, 倘若有重复的直接拼接在后面, 最坏情况下, max1重复很多次, 那要动态开辟一个很大的数组, 在insert时候也无法避免多次比较, 故采用set.

• C++ set
  c++ set是自动排序的!! insert之后 ==> 去重+默认从小到大排序

int thirdMax(vector<int>& nums) {
    set<int> top3;
    for (int num : nums) {
        top3.insert(num);
        if (top3.size() > 3)
            top3.erase(top3.begin());
    }
    return top3.size() == 3 ? *top3.begin() : *top3.rbegin();
}

• Python set
  直观的方式

def thirdMax(self, nums):
        nums = set(nums)
        if len(nums) < 3:
            return max(nums)
        nums.remove(max(nums))
        nums.remove(max(nums))
        return max(nums)

Reference

• leetcode 414. Third Maximum Number