LeetCode 414 Third Maximum Number（set + priority_queue）

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: 1Explanation: The third maximum is 1.

Example 2:

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

Example 3:

Input: [2, 2, 3, 1]Output: 1Explanation: Note that the third maximum here means the third maximum distinct number.Both numbers with value 2 are both considered as second maximum.

题目大意：给出一个非空的整型数组，编写程序返回第三大的数字。如果没有第三大的数字则返回最大的数字。

解题思路：因为数字有可能重复，所以用set维护已经出现的数字，然后再用优先队列来维护数组最大值和第三大的值。

代码如下：

class Solution {public:    int thirdMax(vector<int>& nums) {        for(auto a : nums){            if(st.count(a) == 0){                pque.push(a);                st.insert(a);            }        }        int thirdMax = pque.top();        if((int)st.size() >= 3) {            pque.pop();            pque.pop();            thirdMax = pque.top();        }        return thirdMax;    }private:    priority_queue<int> pque;    set<int> st;};