Third Maximum Number——LeetCode⑬

//原题链接https://leetcode.com/problems/third-maximum-number/

• 题目描述

  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.
  

• 思路分析
  给一个非空的整数数组，返回第三大的数字，不足三个返回最大的数字；
  时间复杂度必须为O(n)
  比较简单，只要维持一个大小3的优先队列不断更新即可
• 源码附录

  class Solution {
      public int thirdMax(int[] nums) {
          if(nums==null || nums.length==0)
          {
              return 0;
          }
          
          PriorityQueue<Integer> priqueue = new PriorityQueue<Integer>();
          Set<Integer> set = new HashSet<Integer>();
          
          for(int i=0;i<nums.length;i++)
          {
              if(!priqueue.contains(nums[i]))
              {
                  priqueue.add(nums[i]);
                  set.add(nums[i]);
                  if(priqueue.size()>3)
                  {
                      set.remove(priqueue.poll());
                  }
              }            
          }
          if(priqueue.size()<3)
              {
                  while(priqueue.size()>1)
                  {
                      priqueue.poll();
                  }
              }
          return priqueue.peek();
      }
  }