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: 1Explanation: The third maximum is 1.

Example 2:

Input: [1, 2]Output: 2Explanation: The third maximum does not exist, so the maximum (2) is returned instead.

Example 3:

Input: [2, 2, 3, 1]Output: 1Explanation: Note that the third maximum here means the third maximum distinct number.Both numbers with value 2 are both considered as second maximum.

找出数组中第三大的数，如果不存在的话那么就返回最大的数。

思路：用三个变量first, second, third来分别保存第一大、第二大和第三大的数，然后遍历数组，如果遍历到的数字大于当前first，那么三个变量各自错位赋值，如果当前数字大于second且小于first，那么就更新second和third，如果当前数字大于third且小于second，那就只更新third。

注意：1初始化要用长整型long的最小值Long.MIN_VALUE，因为当数组中存在整型MIN_VALUE时，无法判断返回第一大还是第三大。当然，可以设置标志位flag=False，如果存在赋值动作就flag=true。

2 最后要换回int型，在前面加(int)third，而不是int(third)。

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