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.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; } }