Third Maximum Number

题目：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.

solution1:维护三个变量，分别保存最大数，次大数和第三大的数。遍历数组不断更新这三个数。遇到的问题，三个变量应该初始化成多少。

solution2：用STL中的set容器，自动去重和排序功能。（但是不知道为啥下面这个代码没有通过编译）

class Solution {public:    int thirdMax(vector<int>& nums) {        int n=nums.size();        set<int,greater<int>> s;        for(int i=0;i<n;i++)        {            s.insert(nums[i]);                    }        set<int>::iterator iter=s.begin();        if(s.size()>=3);        {            iter--;            iter--;        }        return *iter;           }};