LeetCode 414. 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.

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<int>& nums) {vector<int> sort;//保存nums排序后的vectorsort.push_back(nums[0]);//存入nums第一个数for (int i = 1; i<nums.size(); ++i){//将nums中的数遍历存入sort中for (int j = 0; j<sort.size(); ++j){if (sort[j] == nums[i]){//若相等，即重复，break不执行break;}if (sort[j]>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]<nums[i] && j == sort.size() - 1){//如果比所有的数都大，则在末尾push_backsort.push_back(nums[i]);}}}if (sort.size()<3){return *(sort.end() - 1);//sort长度小于3，即没有第三大的数，返回最大的}else{return *(sort.end() - 3);//有第三大的，返回第三大的}}
};

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