链表题--01----K 个一组翻转链表

Posted 2021-12-17 高高for 循环

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文章目录

• K 个一组翻转链表
• • 题目：
  • 分析:
  • 数K个数,返回
  • 链表反转
  • 总逻辑
• 代码

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K 个一组翻转链表

https://leetcode.com/problems/reverse-nodes-in-k-group/

题目：

分析:

数K个数,返回

	public static ListNode getKGroupEnd(ListNode start, int k) 
		while (--k != 0 && start != null) 
			start = start.next;
		
		return start;
	

链表反转

	public static void reverse(ListNode start, ListNode end) 
		end = end.next;
		ListNode pre = null;
		ListNode cur = start;
		ListNode next = null;
		while (cur != end) 
			next = cur.next;
			cur.next = pre;
			pre = cur;
			cur = next;
		
		start.next = end;
	

总逻辑

代码

// 测试链接：https://leetcode.com/problems/reverse-nodes-in-k-group/
public class Code04_ReverseNodesInKGroup 

	// 不要提交这个类
	public static class ListNode 
		public int val;
		public ListNode next;
	

	public static ListNode reverseKGroup(ListNode head, int k) 
		ListNode start = head;
		ListNode end = getKGroupEnd(start, k);
		if (end == null) 
			return head;
		
		// 第一组凑齐了！
		head = end;
		reverse(start, end);
		// 上一组的结尾节点
		ListNode lastEnd = start;
		while (lastEnd.next != null) 
			start = lastEnd.next;
			end = getKGroupEnd(start, k);
			if (end == null) 
				return head;
			
			reverse(start, end);
			lastEnd.next = end;
			lastEnd = start;
		
		return head;
	

	public static ListNode getKGroupEnd(ListNode start, int k) 
		while (--k != 0 && start != null) 
			start = start.next;
		
		return start;
	

	public static void reverse(ListNode start, ListNode end) 
		end = end.next;
		ListNode pre = null;
		ListNode cur = start;
		ListNode next = null;
		while (cur != end) 
			next = cur.next;
			cur.next = pre;
			pre = cur;
			cur = next;
		
		start.next = end;