K 个一组翻转链表
题目介绍
K 个一组翻转链表
给你一个链表，每 k 个节点一组进行翻转，请你返回翻转后的链表。k 是一个正整数，它的值小于或等于链表的长度。如果节点总数不是 k 的整数倍，那么请将最后剩余的节点保持原有顺序。
示例 1：
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| 给你这个链表：1->2->3->4->5
当 k = 2 时，应当返回: 2->1->4->3->5
当 k = 3 时，应当返回: 3->2->1->4->5
|
说明：
- 你的算法只能使用常数的额外空间。
- 你不能只是单纯的改变节点内部的值，而是需要实际进行节点交换。
题目解法
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| package algorithm;
public class ReverseNodesInKGroup {
public static ListNode reverseKGroup(ListNode head, int k) {
if (head == null || head.next == null || k <= 1) {
return head;
}
ListNode prev = head;
ListNode cur = head;
int length = 0;
while (length < k && cur != null) {
length++;
prev = cur;
cur = cur.next;
}
if (length < k){
return head;
}
prev.next = null;
ListNode left = reverse(head, k);
ListNode right = reverseKGroup(cur, k);
left.next = right;
return prev;
}
private static ListNode reverse(ListNode head, int k) {
if (head == null || head.next == null) {
return head;
}
ListNode prev = head;
int length = 0;
while (prev != null) {
length++;
prev = prev.next;
}
if (length < k){
return head;
}
prev = head;
ListNode cur = prev.next;
while (cur != null) {
ListNode next = cur.next;
cur.next = prev;
prev = cur;
cur = next;
}
head.next = null;
return head;
}
public static void main(String[] args) {
ListNode n1 = new ListNode(1);
ListNode n2 = new ListNode(2);
n1.next = n2;
ListNode n3 = new ListNode(3);
n2.next = n3;
ListNode n4 = new ListNode(4);
n3.next = n4;
ListNode n5 = new ListNode(5);
n4.next = n5;
print(reverseKGroup(n1, 3));
n1 = new ListNode(1);
n2 = new ListNode(2);
n1.next = n2;
n3 = new ListNode(3);
n2.next = n3;
n4 = new ListNode(4);
n3.next = n4;
n5 = new ListNode(5);
n4.next = n5;
print(reverseKGroup(n1, 2));
}
private static void print(ListNode head) {
while (head != null) {
System.out.print(head.val);
head = head.next;
}
System.out.println();
}
public static class ListNode {
int val;
ListNode next;
ListNode() {
}
ListNode(int val) {
this.val = val;
}
ListNode(int val, ListNode next) {
this.val = val;
this.next = next;
}
}
}
|
打印：
思路：
这道题解法很简单，先分段反转，然后通过递归拼接。难点在于链表的指针，一不小心就指错了。