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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;
        }        
     
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  • 原文地址:https://www.cnblogs.com/kiplove/p/6986188.html
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