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(leetcode题解)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.

题意寻找一个数组中第三大的数,如果不存在返回最大的数。
直接的做法就是排序,查找第三个即可,C++实现如下
int thirdMax(vector<int>& nums) {
        sort(nums.begin(),nums.end());
        int count=1;
        for(int i=nums.size()-1;i>0;i--)
        {
            if(nums[i]!=nums[i-1])
                count++;
            if(count==3)
                return nums[--i];
        }
        return nums[nums.size()-1];
    }

 也可以利用set集合的有序性和唯一性,C++实现如下

 int thirdMax(vector<int>& nums) {
        set<int> t_set;
        for(auto &i:nums)
            t_set.insert(i);
        auto r_iter=t_set.rbegin();
        if(t_set.size()<3)
            return *r_iter;
        r_iter++;
        r_iter++;
        return *r_iter;
    }        

 

 

(leetcode题解)Third Maximum Number