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.

 

思路：

首先想到的是用一个map存储数组中的值，作为key，因为map中的key是有序的，于是可以直接拿来用。

int thirdMax(vector
     
      & nums){  map
      
       m;  map
       
        ::iterator it;  for(int i=0;i
        
         first;}
        
       
      
     

又看到大神用的set，随时保持set的size不超过3个，这样更简洁！

int thirdMax(vector
     
      & nums) {    set
      
        top3;    for (int num : nums) {        top3.insert(num);        if (top3.size() > 3)            top3.erase(top3.begin());    }    return top3.size() == 3 ? *top3.begin() : *top3.rbegin();}
      
     

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