#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.

   题解：这道题要找到第三大的数，我用了set，避免同一元素出现多次，定义反向迭代器，之后判断set中有多少个元素，小于三个时直接返回最大值，否则反向迭代就可以找到第三大的值了，时间复杂度是O(n)

class Solution {
public:
    int thirdMax(vector<int>& nums) {
        set<int> num;
        for(int i=0;i<nums.size();i++)
        {
            num.insert(nums[i]);
        }
        set<int>::reverse_iterator it=num.rbegin();
        if(num.size()<3)
           return *it; 
           
        it++;
        it++;
        return *it;
    }
};

Leetcode-414 Third Maximum Number