chapter-four/0400~0499/0414.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 }