414. Third Maximum Number

最新推荐文章于 2024-07-10 23:03:37 发布

无名_1989

最新推荐文章于 2024-07-10 23:03:37 发布

阅读量148

点赞数

本文介绍了一种算法，该算法可以在O(n)的时间复杂度内找到整数数组中的第三大数。如果不存在第三大数，则返回最大数。文章通过示例说明了如何使用一个集合来跟踪前三个最大的不同数值。

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.

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();
}