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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.

思路：给定一个vector，找到vector中第三大的不重复的数，若不存在，则返回最大的数。

代码：

class Solution {public:	int thirdMax(vector
   
    & nums) {		vector
    
      sort;//保存nums排序后的vector		sort.push_back(nums[0]);//存入nums第一个数		for (int i = 1; i
     
      nums[i]){//若nums中的数比sort[j]小，则将nums存入当前位置之前，跳出循环					sort.insert(sort.begin() + j, nums[i]);					break;				}				else if (sort[j]
     
    
   

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