414. Third Maximum Number

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.

自己的笨方法, 粗暴, 易理解, 且符合题目要求:

使用了set.

扫描三遍数组:
第一遍: 找出max, 并依次把每个元素送入set, 若 set.size() < 3, 返回max;
第二遍找出sec, 第三遍找出third;
返回 third.

(O(n)) time, (O(n)) extra space.

int thirdMax(vector<int>& A) {
    int i, n = A.size();
    int max = INT_MIN, sec = INT_MIN, third = INT_MIN;
    set<int> set;

    for (i = 0; i < n; i++) {
        set.insert(A[i]);
        max = max > A[i] ? max : A[i];
    }

    if (set.size() < 3) return max;

    for (i = 0; i < n; i++)
        if (A[i] != max)
            sec = sec > A[i] ? sec : A[i];

    for (i = 0; i < n; i++)
        if (A[i] != max && A[i] != sec)
            third = third > A[i] ? third : A[i];

    return third;
}

人家的精巧方法:
用到set, set.erase(), set.begin(), set.rbegin()
set 默认升序. set.rbegin()表示最后一个元素的迭代器(指针).
(O(n)) time, (O(n)) extra space.

int thirdMax(vector<int>& A) {
    set<int> top3;
    for (int i = 0; i < A.size(); i++) {
        top3.insert(A[i]);
        // set 默认按 key 升序排序
        if (top3.size() > 3) top3.erase(top3.begin());
    }
    return top3.size() == 3 ? *top3.begin() : *top3.rbegin();
}