Shulou(Shulou.com)06/01 Report--

I would like to share with you how java judges the existence of repetitive elements. I hope you will gain something after reading this article. Let's discuss it together.

Given an array of integers, determine whether there are duplicate elements.

If any value appears in the array at least twice, the function returns true. Returns false if each element in the array is different.

Example 1:

Input: [1, 2, 2, 3, 1]

Output: true

Example 2:

Input: [1, 2, 3, 4]

Output: false

Example 3:

Input: [1, 1, 1, 1, 3, 3, 4, 3, 4, 3, 4, 4, 4, 2]

Output: true

The problem in the previous issue is: 157, reverse the linked list

1public ListNode reverseList (ListNode head) {

2 if (head = = null | | head.next = = null)

3 return head

4 ListNode tempList = reverseList (head.next)

5 head.next.next = head

6 head.next = null

7 return tempList

8}

Parsing:

List reversal, this is a clich é problem, in fact, there are many ways, let's take a look at two more

1public ListNode reverseList (ListNode head) {

2 ListNode pre = null

3 while (head! = null) {

4 ListNode next = head.next

5 head.next = pre

6 pre = head

7 head = next

8}

9 return pre

10}

eleven

twelve

13public ListNode reverseList (ListNode head) {

14 return reverseListInt (head, null)

15}

sixteen

17private ListNode reverseListInt (ListNode head, ListNode newHead) {

18 if (head = = null)

19 return newHead

20 ListNode next = head.next

21 head.next = newHead

22 return reverseListInt (next, head)

23} after reading this article, I believe you have some understanding of "how java judges the existence of repetitive elements". If you want to know more about it, you are welcome to follow the industry information channel. Thank you for reading!

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