3. 25_k个一组翻转链表
/*
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
*/
/*方法一*/
class Solution {
public ListNode reverseKGroup(ListNode head, int k) {
ListNode dummy = new ListNode(0), prev = dummy, curr = head, next;
dummy.next = head;
int length = 0;
while(head != null) {
length++;
head = head.next;
}
head = dummy.next;
for(int i = 0; i < length / k; i++) {
for(int j = 0; j < k - 1; j++) {
next = curr.next;
curr.next = next.next;
next.next = prev.next;
prev.next = next;
}
prev = curr;
curr = prev.next;
}
return dummy.next;
}
}
/*方法二*/
public class Solution {
public ListNode reverseKGroup(ListNode head, int k) {
ListNode hair = new ListNode(0);
hair.next = head;
ListNode pre = hair;
ListNode end = hair;
while (end.next != null) {
for (int i = 0; i < k && end != null; i++){
end = end.next;
}
if (end == null){
break;
}
ListNode start = pre.next;
ListNode next = end.next;
end.next = null;
pre.next = reverse(start);
start.next = next;
pre = start;
end = pre;
}
return hair.next;
}
private ListNode reverse(ListNode head) {
ListNode pre = null;
ListNode curr = head;
while (curr != null) {
ListNode next = curr.next;
curr.next = pre;
pre = curr;
curr = next;
}
return pre;
}
}
/*方法三 递归*/
class Solution {
public ListNode reverseKGroup(ListNode head, int k) {
ListNode prev = null;
ListNode cur = head;
ListNode next = null;
ListNode check = head;
int canProceed = 0;
int count = 0;
// 检查链表长度是否满足翻转
while (canProceed < k && check != null) {
check = check.next;
canProceed++;
}
// 满足条件,进行翻转
if (canProceed == k) {
while (count < k && cur != null) {
next = cur.next;
cur.next = prev;
prev = cur;
cur = next;
count++;
}
if (next != null) {
// head 为链表翻转后的尾节点
head.next = reverseKGroup(next, k);
}
// prev 为链表翻转后的头结点
return prev;
} else {
// 不满住翻转条件,直接返回 head 即可
return head;
}
}
}