414. Third Maximum Number

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最新推荐文章于 2024-07-10 23:03:37 发布

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分类专栏： LeetCode LeetCode 文章标签： LeetCode Kth-Number 第K大数 414

本文链接：https://blog.youkuaiyun.com/teffi/article/details/54179236

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本文介绍了一种在O(n)时间复杂度内找到非空整数数组中第三大数值的方法。若不存在第三大数值，则返回最大值。文章通过实例演示了如何通过设置最大、次大及第三大的变量来遍历数组并更新这些变量。

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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.

解题思路：
设置最大、第二大、第三大的变量，遍历数组，更新这三个变量，考虑相同的数字时，变更三个变量的策略。

class Solution {
public:
    int thirdMax(vector<int>& nums) {
        long first_max = LONG_MIN;
        long second_max = LONG_MIN;
        long third_max = LONG_MIN;

        for (auto num : nums) {
            if (num > first_max) {
                if (second_max != LONG_MIN) third_max = second_max;
                if (first_max != LONG_MIN) second_max = first_max;
                first_max = num;
            }
            else if (num > second_max && num != first_max) {
                if (second_max != LONG_MIN) third_max = second_max;
                second_max = num;
            }
            else if (num > third_max && num != second_max && num != first_max) 
                third_max = num;
        }

        return third_max == LONG_MIN ? first_max : third_max;
    }
};