LeetCode414 Third Maximum Number

最新推荐文章于 2024-07-10 23:03:37 发布

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最新推荐文章于 2024-07-10 23:03:37 发布

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本文介绍了一个算法问题：如何在一个整数数组中找到第三大的数，若不存在则返回最大数。该算法采用O(n)的时间复杂度，通过遍历数组并维护三个变量来实现。文章包含示例输入输出及解释。

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题目要求:

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 fix {
    public int thirdMax(int[] nums) {

        //创建三个变量用于记录数值的大小
        long max = Long.MIN_VALUE;
        long mid = Long.MIN_VALUE;
        long sml = Long.MIN_VALUE;
        int count = 0;

        for(int i : nums) {
            if(i > max) {
                sml = mid;
                mid = max;
                max = i;
                count++;
            }
            else if(max > i && i > mid) {
                sml = mid;
                mid = i;
                count++;
            }
            else if(mid > i && i >= sml) {
                sml = i;
                count++;
            }

        }

        if(count >= 3) return (int)sml;
        else return (int)max;
    }
}