Third Maximum Number

Posted 2020-10-12

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.

 取三个Integer，遍历数组

为什么要Integer？因为int默认值为0，不好比较

public int thirdMax(int[] nums) {
          Integer max1 = null;
           Integer max2 = null;
           Integer max3 = null;
           for(Integer n:nums) {
               if(n.equals(max1)||n.equals(max2)||n.equals(max3)) continue;
               if(max1==null||n>max1) {
                   max3=max2;
                   max2=max1;
                   max1=n;
               }
               else if(max2==null||n>max2) {
                   max3=max2;
                   max2=n;
               }
               else if(max3==null||n>max3) {
                   max3=n;
               }
           }
           return max3==null?max1:max3;//如果max用int，int默认值为0且不能为null，就会与数组本身的数冲突，分不清究竟数组中是否有第三大的数
    }