一、题目描述:
二、思路分析:
翻转链表应该是很多人都做过的面试题了,这里给没做过的简单说下
指针pre = null,next = head.next, tmp = head;
循环tmp-> next = pre; pre = tmp;tmp = next;next = tmp->next;
return tmp
而这个k组翻转其实就是先将链表分为:未翻转,正在翻转,已经翻转三部分,循环一遍得到。
三、AC 代码:
/**
* 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 {
public ListNode reverseKGroup(ListNode head, int k) {
ListNode dummy = new ListNode(0);
dummy.next = head;
ListNode pre = dummy;
ListNode end = dummy;
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 dummy.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;
}
}
四、总结:
这是一道困难题,我在会翻转链表的前提下还是没做出来,题解中前驱节点dummy的设置我觉得十分的妙,(虽然这个在翻转链表里也出现过),只是当初只是当模板背的,并没有吃透该知识点
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