LeetCode刷题【Array】Third Maximum Number

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本文介绍了一种在O(n)时间复杂度内找到数组中第三大数的算法，并提供了具体的Java实现代码。当数组中不包含第三个最大数时，算法将返回最大数。

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.

解决方案： Runtime 13ms

public class Solution {
    public int thirdMax(int[] nums) {
        Integer a[] = new Integer[3];
        for(Integer num: nums){
            if(num.equals(a[0])||num.equals(a[1])||num.equals(a[2])) continue;
            if(a[0]==null||num>a[0]) {a[2]=a[1];a[1]=a[0];a[0]=num;}
            else if(a[1]==null||num>a[1]){a[2]=a[1];a[1]=num;}
            else if(a[2]==null||num>a[2]){a[2]=num;}
        }
        
        return a[2]==null? a[0]:a[2];
    }
}

参考：

【1】https://leetcode.com/