LeetCode - 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.
class Solution { public int thirdMax(int[] nums) { long first = Long.MIN_VALUE; long second = Long.MIN_VALUE; long third = Long.MIN_VALUE; for (int n : nums) { if (n > first) { third = second; second = first; first = n; } else if (n < first && n > second) { third = second; second = n; } else if (n < second && n > third) { third = n; } } if ( third == Long.MIN_VALUE ) { return (int)first; } return (int)third; } }
posted on 2020-01-16 14:26 IncredibleThings 阅读(117) 评论(0) 收藏 举报