25. K个一组翻转链表
查看题目困难
链表
递归
频率: 475
解法一:迭代法
时间复杂度:O(n) | 空间复杂度:O(1) | 推荐使用
动画演示
准备就绪 - 输入链表值(用逗号分隔)和k值,然后点击开始
代码实现
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;
}
时间复杂度:O(n)
空间复杂度:O(1)
使用迭代法,每次反转k个节点,然后移动到下一个k个节点组。