25. Reverse Nodes in k Group - jiejackyzhang/leetcode-note GitHub Wiki

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

If the number of nodes is not a multiple of k then left-out nodes in the end should remain as it is.

You may not alter the values in the nodes, only nodes itself may be changed.

Only constant memory is allowed.

For 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

Linked List类题目。

与前一道题类似。注意点为先确定一组k-group，然后将之reverse。可以作递归处理。

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
public class Solution {
    public ListNode reverseKGroup(ListNode head, int k) {
        if(head == null || k <= 1) {
            return head;
        }
        ListNode curr = head;
        int count = 0;
        while(curr != null && count < k) {
            // find k+1 node
            curr = curr.next;
            count++;
        }
        if(count == k) {
            curr = reverseKGroup(curr, k);
            //reserve current k group nodes
            while(count > 0) {
                ListNode temp = head.next;
                head.next = curr;
                curr = head;
                head = temp;
                count--;
            }
            head = curr;
        }
        return head;
    }
}