LeetCode 414. Third Maximum Number

414. Third Maximum Number

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• Total Accepted: 22549
• Total Submissions: 83903
• Difficulty: Easy
• Contributors: ZengRed , 1337c0d3r

 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.

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【题目分析】

给定一个整数数组，返回数组中第三大的数，如果该数不存在，则返回最大的那个数。要求时间复杂度为O(n).

【思路分析】

1. 找到最大的数和最小的数

2. 遍历数组，找到次小的数

3. 如果次小的数存在，再次遍历数组找到第三小的数。

【java代码】

public class Solution {
    public int thirdMax(int[] nums) {
        if(nums.length == 1) return nums[0];
        if(nums.length == 2) 
            return nums[0] > nums[1] ? nums[0] : nums[1];
        
        int firstmax = nums[0];
        int firstmin = nums[0];
        for(int i = 1; i < nums.length; i++) {
            firstmax = Math.max(firstmax, nums[i]);
            firstmin = Math.min(firstmin, nums[i]);
        }
        
        int secondmax = firstmin;
        for(int i = 0; i < nums.length; i++) {
            if(nums[i] < firstmax)
                secondmax = Math.max(secondmax, nums[i]);
        }
        
        if(secondmax < firstmax) {
            int thirdmax = firstmin;
            for(int i = 0; i < nums.length; i++) {
                if(nums[i] < secondmax)
                    thirdmax = Math.max(thirdmax, nums[i]);
            }
            if(thirdmax < secondmax) return thirdmax;
        }
        return firstmax;
    }
}

上述代码的缺点是每次遍历时要先确定一个下届，因此必须先找到数组中最小的那个值。但是算法的时间效率还是不错的。一个更加简明的程序如下：

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;
    }