414. 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.

法1：使用快排中的“切分” O(NlgN)

public class Solution {
    public int thirdMax(int[] nums)
	{
        
		HashSet<Integer> hashset=new HashSet<>(nums.length);
        for(int n : nums)
        	hashset.add(n);
        
        int[] arr=new int[hashset.size()];
        int cnt=0;
        for(int n :hashset)
        	arr[cnt++]=n;
        
        int len=arr.length;
        
        if(len<3)
        {
        	int max=Integer.MIN_VALUE;
        	for(int i=0;i<len;i++)
        		max=Math.max(arr[i], max);
        	return max;
        }
        
       
        
       return findkth(arr, len-3, 0, len-1);
        
    }
	
	public int findkth(int[] nums,int k,int lo,int hi)
	{
		int kth=partition(nums, lo, hi);
		int ele=nums[kth];
		
		if(kth>k)
			return findkth(nums, k, lo, kth);
		else if(kth<k)
			return findkth(nums, k, kth+1, hi);
		else {
			return ele;
		}
	}
	
	public int partition(int[] nums,int lo,int hi)
	{
		if(lo==hi)
			return lo;
		
		int num=nums[lo];
		int i=lo,j=hi+1;
		while(true)
		{
			while(nums[++i]<num)
				if(i==hi)
					break;
			while(nums[--j]>num)
				if(j==lo)
					break;
			if(i>=j)
				break;
			
			swap(nums, i, j);
		}
		
		swap(nums, lo, j);
		
		return j;
	}
	public static void swap(int[] arr ,int i, int j)
	{
		int temp=arr[i];
		arr[i]=arr[j];
		arr[j]=temp;
	}
}

法2 ，使用三个变量直接检索，摘自

https://discuss.leetcode.com/topic/62236/java-solution-in-0ms-run-time-o-n-and-space-o-1/15

public int thirdMax(int[] nums) {
    long max = Long.MIN_VALUE, mid = max, min = max;
        
    for (int ele : nums) {
        if (ele > max) {
            min = mid;
            mid = max;
            max = ele;
        } else if (max > ele && ele > mid) {
            min = mid;
            mid = ele;
        } else if (mid > ele && ele > min) {
            min = ele;
        }
    }
        
    return (int)(min != Long.MIN_VALUE ? min : max);
}