leetcode414. Third Maximum Number

最新推荐文章于 2025-08-22 19:21:44 发布

原创最新推荐文章于 2025-08-22 19:21:44 发布 · 223 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#leetcode #array

leetcode 专栏收录该内容

187 篇文章

订阅专栏

本文介绍了一种高效算法，用于从整数数组中找出第三大的数。若不存在，则返回最大值。文章提供了两种解决方案：一种使用三个变量维护前三大的数；另一种借助优先队列和集合实现。这两种方法均满足O(n)的时间复杂度要求。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

414. Third Maximum Number

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.

解法一

维护3个数，max1,max2,max3，分别存放前3大数，为什么不取int max1 = Integer.Integer.MIN_VALUE,因为
if (max3 > Integer.MIN_VALUE) {
return max3;
}
如果数组中有3个数{0，1，Integer.MIN_VALUE}，即不通过。

public class Solution {
    public int thirdMax(int[] nums) {
        if (nums == null || nums.length == 0) {
            return -1;
        }

        Integer max1 = null;
        Integer max2 = null;
        Integer max3 = null;
        for (Integer num : nums) {
            if (num.equals(max1) || num.equals(max2) || num.equals(max3))
                continue;
            if (max1 == null || num > max1) {
                max3 = max2;
                max2 = max1;
                max1 = num;
            } else if (max2 == null || num > max2) {
                max3 = max2;
                max2 = num;
            } else if (max3 == null || num > max3) {
                max3 = num;
            }
        }

        if (max3 != null) {
            return max3;
        }
        return max1;
    }
}

解法二

利用优先队列和set，优先队列只保留3位，所有peek元素就是第3大元素。如果队列的长度小于3，队列中只保留1位元素，即最大的元素。

public class Solution {
    public int thirdMax(int[] nums) {
        if (nums == null || nums.length == 0) {
            return -1;
        }

        PriorityQueue<Integer> pq = new PriorityQueue<>();
        Set<Integer> set = new HashSet<>();
        for (int num : nums) {
            if (!set.contains(num)) {
                pq.offer(num);
                set.add(num);
                if (pq.size() > 3) {
                    set.remove(pq.poll());
                }
            }
        }
        if (pq.size() < 3) {
            while (pq.size() > 1) {
                pq.poll();
            }
        }

        return pq.peek();
    }
}