[算法] 将单链表的每K个节点之间逆序

题目

给定一个单链表的头结点，实现一个调整单链表的函数，使得每K个节点之间逆序，如果最后不够K个节点一组，则不调整最后几个节点。

解答

使用栈结构

import java.util.Stack;
 public class Test{
 static class Node{
	public int val;
	public Node next;
	public Node(int val){
		this.val=val;
	}
}
public static void main(String[] args) {
	Node head=new Node(1);
	head.next=new Node(2);
	head.next.next=new Node(3);
	head.next.next.next=new Node(4);
	head.next.next.next.next=new Node(5);
	head.next.next.next.next.next=new Node(6);
	Node node=reverseKNodes(head,2);
	while(node!=null){
		System.out.print(node.val+" ");
		node=node.next;
	}
}

public static Node reverseKNodes(Node head, int K){
	if (K<2) {
		return head;
	}
	Stack<Node> stack=new Stack<>();
	Node newHead=head;
	Node cur=head;
	Node pre=null;
	Node next=null;
	while(cur!=null){
		stack.push(cur);
		next=cur.next;
		if (stack.size()==K) {
			pre=resign(stack,pre,next);
			newHead=newHead==head?cur:newHead;
		}
		cur=next;
	}
	return newHead;
}
public static Node resign(Stack<Node> stack,Node left,Node right){
		Node cur=stack.pop();
		if (left!=null) {
			left.next=cur;
		}
		Node next=null;
		while(!stack.isEmpty()){
			next=stack.pop();
			cur.next=next;
			cur=next;
		}
		cur.next=right;
		return cur;
}
 }

输出：2 1 4 3 6 5

不使用栈结构

import java.util.Stack;
public class Test{
 static class Node{
	public int val;
	public Node next;
	public Node(int val){
		this.val=val;
	}
}
public static void main(String[] args) {
	Node head=new Node(1);
	head.next=new Node(2);
	head.next.next=new Node(3);
	head.next.next.next=new Node(4);
	head.next.next.next.next=new Node(5);
	head.next.next.next.next.next=new Node(6);
	Node node=reverseKNodes(head,2);
	while(node!=null){
		System.out.print(node.val+" ");
		node=node.next;
	}
}

public static Node reverseKNodes(Node head, int K){
	if (K<2) {
		return head;
	}
	Node cur=head;
	Node pre=null;
	Node next=null;
	Node start=null;
	int count=1;
	while(cur!=null){
		next=cur.next;
		if (count==K) {
			start=pre==null?head:pre.next;
			head=pre==null?cur:head;
			resign(pre,start,cur,next);
			pre=start;
			count=0;
		}
		count++;
		cur=next;
	}
	return head;
}
public static void resign(Node left,Node start,Node end, Node right){
	Node pre=start;
	Node cur=start.next;
	Node next=null;
	while(cur!=right){
		next=cur.next;
		cur.next=pre;
		pre=cur;
		cur=next;
	}
	if (left!=null) {
		left.next=end;
	}
	start.next=right;
}

}