414. Third Maximum Number

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最新推荐文章于 2024-07-10 23:03:37 发布

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博客提出一个算法问题，给定非空整数数组，需返回数组中第三大的数字，若不存在则返回最大数字，且要求时间复杂度为O(n)，还给出了示例，但未给出具体思路。

Given a non-empty array of integers, return the thirdmaximum 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.

思路参考这里

class Solution {
public:
    int thirdMax(vector<int>& nums) {
        long first=LONG_MIN,second=LONG_MIN,third=LONG_MIN;
        for(int i=0;i<nums.size();i++){
            if(nums[i]>first){
                third=second;
                second=first;
                first=nums[i];
            }
            else if(nums[i]<first&&nums[i]>second){
                third=second;
                second=nums[i];
            }
            else if(nums[i]<second&&nums[i]>third){
                third=nums[i];
            }
        }
        return (third==second||third==LONG_MIN)?first:third;
    }
};