0414. Third Maximum Number

414. Third Maximum Number #

题目 #

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.

题目大意 #

给定一个非空数组,返回此数组中第三大的数。如果不存在,则返回数组中最大的数。要求算法时间复杂度必须是 O(n)。

解题思路 #

  • 水题,动态维护 3 个最大值即可。注意数组中有重复数据的情况。如果只有 2 个数或者 1 个数,则返回 2 个数中的最大值即可。

代码 #


package leetcode

import (
	"math"
)

func thirdMax(nums []int) int {
	a, b, c := math.MinInt64, math.MinInt64, math.MinInt64
	for _, v := range nums {
		if v > a {
			c = b
			b = a
			a = v
		} else if v < a && v > b {
			c = b
			b = v
		} else if v < b && v > c {
			c = v
		}
	}
	if c == math.MinInt64 {
		return a
	}
	return c
}


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