Array-Third Maximum Number

Description:
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.

Solution:

class Solution {
      public int thirdMax(int[] nums) {
        Integer max1 = null;
        Integer max2 = null;
        Integer max3 = null;
        //注意不同元素
        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;
    }
}

Best Solution:

class Solution {
    public int thirdMax(int[] nums) {
        long first = Long.MIN_VALUE;
        long second =Long.MIN_VALUE;
        long third = Long.MIN_VALUE;

        for(int i = 0; i< nums.length; i++)
        {
            if(nums[i] > first)
            {
                third = second;
                second = first;
                first = nums[i];
            }else if(nums[i] > second && nums[i] < first)
            {
                third = second;
                second = nums[i];
            }else if( nums[i] > third && nums[i] < second)
            {
                third = nums[i];
            }
        }
        return third == Long.MIN_VALUE ? (int)first : (int)third;
    }
}

总结:设置3个数,遍历数组划分数据，需要注意的是每次添加的是不同的数据。