（leetcode题解）Third Maximum Number

标签：数组   output   利用   rdm   imu   sid   rdma   排序   return   

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

标签：数组   output   利用   rdm   imu   sid   rdma   排序   return   

原文地址：http://www.cnblogs.com/kiplove/p/6986188.html