414.Third Maximum Number

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.


class Solution {
    int thirdMax(vector<int>& nums) {
        if(nums.size()<3) return *max_element(nums.begin(),nums.end());

        for(int i=0;i<nums.size();i++)
            if(ele.count(nums[i])) continue;
        if(three.size()==1) return three.top();
        if(three.size()==2)    {three.pop();    return three.top();}
        while(three.size() !=3) three.pop();
        return three.top();

执行用时 :24 ms, 在所有 C++ 提交中击败了13.20%的用户 内存消耗 :10.8 MB, 在所有 C++ 提交中击败了5.05%的用户

Last updated