一、問題描述

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).

Example1

Input: [3, 2, 1]

Output: 1

Explanation: The third maximum is 1.

Example2

Input: [1, 2]

Output: 2

Explanation: The third maximum does not exist, so the maximum (2) is returned instead.
 

Example3

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.

二、題意分析

找到第三大的數，如果沒有第三大的數就返回最大的數

三、解題思路

直接對第一，第二，第三大的數分別進行判斷，要注意的是陣列中的數會取到整數的最小值。

程式碼實現

class Solution
 
 {
    public int thirdMax(int[] nums) {
        if(nums.length == 1)
            return nums[0];
        else if(nums.length == 2)
            return Math.max(nums[0], nums[1]);
        int max = nums[0];
        for(int i = 1; i < nums.length; i++){
            max = Math.max(max, nums[i]);     
        }
 


        //arr儲存第一，第二，第三大數
        int[] arr = new int[3];
        arr[0] = Integer.MIN_VALUE;
        arr[1] = Integer.MIN_VALUE;
        arr[2] = max;
        int counts = 1;
        boolean flag = false;
        for(int key : nums){
            // 處理整數最小值的情況
            if(key == Integer.MIN_VALUE)
                flag = true;
            if(key <= arr[0])
                continue;
            // 第二大數
            if(key < arr[2] && key > arr[1]){
                int tmp = arr[1];
                 arr[1] = key;
                 arr[0] = tmp;
                counts++;
            }
            // 第三大數
            if(key > arr[0] && key < arr[1]){
                arr[0] = key;
                counts++;
            }
               
        }
        if(counts > 2)
            return arr[0];
        else if(flag && counts == 2)
            return Integer.MIN_VALUE;
        return max;  
    }
}

時間、空間複雜度分析

1. 時間複雜度

O (n)

2. 空間複雜度

O（1）