LeetCode25 Reverse Nodes in K Groups

Posted on In

题目描述

Given a linked list, reverse the nodes of a linked list k at a time and return its modified list.

k is a positive integer and is less than or equal to the length of the linked list. If the number of nodes is not a multiple of k then left-out nodes in the end should remain as it is.

Example:

Given this linked list: 1->2->3->4->5

For k = 2, you should return: 2->1->4->3->5

For k = 3, you should return: 3->2->1->4->5

Note:

  • Only constant extra memory is allowed.
  • You may not alter the values in the list’s nodes, only nodes itself may be changed.

    解析

    反转链表的进化版本，每K个分成一组进行反转，最后再连起来，递归可解，每次返回反转后的链表头，注意循环条件，如果找不到K个不反转，直接返回原来的序列，代码如下:
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
public ListNode reverseKGroup(ListNode head, int k) {
    ListNode cur = head;
    ListNode start = head;
    int count = 1;
    while(cur != null){
        cur = cur.next;
        count++;
        if(count == k && cur != null){
            ListNode prev = reverseKGroup(cur.next, k);
            head = cur;
            cur = start;
            for(int i = 0; i < k; i++){
                ListNode tmp = cur.next;
                cur.next = prev;
                prev = cur;
                cur = tmp;
            }
        }
    }
    return head;
}

总结

十分常见的链表题，leetcode 25题老题目了，虽然是hard但是熟记规律也不难掌握，可以多写几遍，这样递归以及反转链表就能掌握得比较透彻了，如果面试问到这题而不会的话是很不应该的。