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 public class ThirdMaximumNumber { public int thirdMax(int[] nums) { //这里要使用long，因为最后返回的是first还是third是根据这个判断的 //也可以维护一个count变量 count++ 如果大于3 就返回third long first=Long.MIN_VALUE,second=Long.MIN_VALUE,third=Long.MIN_VALUE; for(int i=0;i<nums.length;i++) { //去除重复值的影响 if(nums[i]==first || nums[i]==second || nums[i]==third) continue; if(nums[i]>first) { third = second; second = first; first = nums[i]; } else if(nums[i]>second) { third = second; second = nums[i]; } else if(nums[i]>third){ third = nums[i]; } } if(third==Long.MIN_VALUE) { return (int)first; } else { return (int)third; } } public int thirdMax2(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; } }