题目:https://leetcode-cn.com/problems/reverse-nodes-in-k-group/
首先看对应的解题思(算法小抄),看明白思路,自己独立的编写一遍:
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode() {}
* ListNode(int val) { this.val = val; }
* ListNode(int val, ListNode next) { this.val = val; this.next = next; }
* }
*/
class Solution {
//递归问题——返回k个一组的反转链表的头结点
public ListNode reverseKGroup(ListNode head, int k) {
//边界处理
if(head==null){
return null;
}
ListNode a=head,b = head;
//思路,反转前k个结点,返回反转部分的‘下一个结点’
for(int i=0;i<k;i++){
if(b==null){
//不够k个,不用处理了
return head;//返回对应的头结点
}
b = b.next;
}
//不包含b
ListNode newHead = reverseAandB(a,b);//a~b反转以后就变为b,a
a.next = reverseKGroup(b,k);
return newHead;
//然后以反转之后,’下一个结点‘为起点继续反转
}
//反转区间为[a,b)的结点,返回的是反转区间的头结点
public ListNode reverseAandB(ListNode a,ListNode b){
ListNode pre, cur, next;
pre = null;cur=a;next=a;
while(cur != b){
//逐个结点反转
next = cur.next;
cur.next=pre;
pre = cur;
//更新cur
cur = next;
}
//返回反转之后的头结点
return pre;
}
}