leetcode414. Third Maximum Number

最新推荐文章于 2024-07-10 23:03:37 发布

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本文介绍了一道编程题的解决方法，题目要求从整数数组中找出第三大的数，若不存在则返回最大数。通过两个不同的解决方案展示了如何高效地解决问题，并讨论了代码实现细节。

题目

easy程度题

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.

第一次做题，一个easy的题都写了一个多小时- -！

我的代码，先来个选择排序，然后，if，if，if- -

class Solution
{
public:
    int thirdMax(vector<int>& a)
    {
        int n=a.size();
        if(n==1)
            return a[0];
        
        if(n==2)
            return a[0]>=a[1]?a[0]:a[1];
        
        bool sorted= false;
        for(int size=n;(!sorted)&&(size>1);size--)
    	{
    		sorted=true;
    		int indexOfMax=0;
    		for(int i=1;i<size;i++)
    			if(a[indexOfMax]<=a[i]) indexOfMax=i;
    			else  sorted=false;
    		exchange(a[indexOfMax],a[size-1]);
    	}
        
        int p=0,j,result;
        for(j=n-1;j>0;j--)
        {
            
            if(a[j]>a[j-1]) ++p;
            if(p==2)
            {
                result=a[j-1];
                break;
            }
            
            if(j==1&&p==1)
            {   
                result=a[n-1];
                break;
            }
            
            if(j==1&&p==0)
            {
                result=a[n-1];
                break;
            }
        }
        
        return result;
    }
    
    void exchange(int &a, int &b)
    {
    	int temp=a;
    	a=b;
    	b=temp;
    }
};

高手写的代码

public class Solution  
{  
    public int thirdMax(int[] nums)  
    {  
        long first, second, third;  
        first = second = third = Long.MIN_VALUE;  
        for (int num : nums)  
        {  
            if (num == first || num == second || num == third) continue;  
            if (num > first)  
            {  
                third = second;  
                second = first;  
                first = num;  
            }  
            else if (num > second)  
            {  
                third = second;  
                second = num;  
            }  
            else if (num > third) third = num;  
        }  
  
        return (third == Long.MIN_VALUE) ? (int)first : (int)third;  
    }  
}  

哎，还是太年轻了！