Reverse Nodes in k-Group

最新推荐文章于 2022-04-08 22:28:12 发布

最新推荐文章于 2022-04-08 22:28:12 发布 · 88 阅读

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本文介绍了一种算法，该算法可以将链表中的节点每K个一组进行反转，并返回修改后的链表。该算法不会改变节点中的值，只改变节点本身。此外，如果节点数量不是K的倍数，则剩余的节点保持原样。

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

/**
 * 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 (k < 2) {
        	return head;
        }
        ListNode cur = head;
        ListNode start = null;
        ListNode pre = null;
        ListNode next = null;
        int cnt = 1;
        while (cur != null) {
        	next = cur.next;
        	if (cnt == k) {
        		start = pre == null ? head : pre.next;
        		head = pre == null ? cur : head;
        		resign(pre, start, cur, next);
        		pre = start;
        		cnt = 0;
        	}
        	cnt++;
        	cur = next;
        }
        return head;
    }

	private void resign(ListNode left, ListNode start, ListNode end,
			ListNode right) {
		// TODO Auto-generated method stub
		ListNode pre = start;
		ListNode cur = start.next;
		ListNode next = null;
		while (cur != right) {
			next = cur.next;
			cur.next = pre;
			pre = cur;
			cur = next;
		}
		if (left != null) {
			left.next = end;
		}
		start.next = right;
	}
}