给定一个单链表的头节点head,实现一个调整单链表的函数,使得每K个节点之间逆 该过序,如果最后不够K个节点一组,则不调整最后几个节点。例如:
链表: 1>2>3>4>5>6>7>8>null K=3。
调整后为: 3-2-1-6-5-4-7-8>null 其中7, 8不调整,因为不够一组。
import java.util.Stack;
public class Problem_12_ConvertEveryKNodesInList {
public static class Node {
public int value;
public Node next;
public Node(int data) {
this.value = data;
}
}
public static Node reverseKNodes1(Node head, int K) { //利用栈结构
if (K < 2) {
return head;
}
Stack<Node> stack = new Stack<Node>();
Node newHead = head;
Node cur = head;
Node pre = null;
Node next = null;
while (cur != null) {
next = cur.next;
stack.push(cur);
if (stack.size() == K) {
pre = resign1(stack, pre, next);
newHead = newHead == head ? cur : newHead;
}
cur = next;
}
return newHead;
}
public static Node resign1(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;
}
public static Node reverseKNodes2(Node head, int K) { //直接在原链表修改
if (K < 2) {
return head;
}
Node cur = head;
Node start = null;
Node pre = null;
Node next = null;
int count = 1;
while (cur != null) {
next = cur.next;
if (count == K) {
start = pre == null ? head : pre.next;
head = pre == null ? cur : head;
resign2(pre, start, cur, next);
pre = start;
count = 0;
}
count++;
cur = next;
}
return head;
}
public static void resign2(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;
}
public static void printLinkedList(Node head) {
System.out.print("Linked List: ");
while (head != null) {
System.out.print(head.value + " ");
head = head.next;
}
System.out.println();
}
public static void main(String[] args) {
Node head = null;
int K = 3;
printLinkedList(head);
head = reverseKNodes1(head, K);
printLinkedList(head);
head = reverseKNodes2(head, K);
printLinkedList(head);
System.out.println("=======================");
head = new Node(1);
K = 3;
printLinkedList(head);
head = reverseKNodes1(head, K);
printLinkedList(head);
head = reverseKNodes2(head, K);
printLinkedList(head);
System.out.println("=======================");
head = new Node(1);
head.next = new Node(2);
K = 2;
printLinkedList(head);
head = reverseKNodes1(head, K);
printLinkedList(head);
head = reverseKNodes2(head, K);
printLinkedList(head);
System.out.println("=======================");
head = new Node(1);
head.next = new Node(2);
K = 3;
printLinkedList(head);
head = reverseKNodes1(head, K);
printLinkedList(head);
head = reverseKNodes2(head, K);
printLinkedList(head);
System.out.println("=======================");
head = new Node(1);
head.next = new Node(2);
head.next.next = new Node(3);
head.next.next.next = new Node(4);
K = 2;
printLinkedList(head);
head = reverseKNodes1(head, K);
printLinkedList(head);
head = reverseKNodes2(head, K);
printLinkedList(head);
System.out.println("=======================");
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);
head.next.next.next.next.next.next = new Node(7);
head.next.next.next.next.next.next.next = new Node(8);
K = 3;
printLinkedList(head);
head = reverseKNodes1(head, K);
printLinkedList(head);
head = reverseKNodes2(head, K);
printLinkedList(head);
System.out.println("=======================");
}
运行结果: