Given a linked list, reverse the nodes of a linked list k at a time and return its modified list.

k is a positive integer and is less than or equal to the length of the linked list. If the number of nodes is not a multiple of k then left-out nodes in the end should remain as it is.

Example:

Given this linked list: 1->2->3->4->5

For k = 2, you should return: 2->1->4->3->5

For k = 3, you should return: 3->2->1->4->5

Note:

  • Only constant extra memory is allowed.
  • You may not alter the values in the list's nodes, only nodes itself may be changed.

题意:

给定一个链表,每k个节点一组,做一次反转。

Solution1:we can still simplify this question into how to reverse a linked list, the only difference is we need to set "dummy" and "null" like the left and right boundary by ourselves.

(1) set a pointer pre as a " dummy " ahead

(2) set a pointer last as a " null " boundary

(3) iteratively move cur into the front(pre.next)

(4) cur = next

(5)iteratively move cur into the front(pre.next) until cur meet the "null" boundary

[leetcode]25. Reverse Nodes in k-Group每k个节点反转一下-LMLPHP

[leetcode]25. Reverse Nodes in k-Group每k个节点反转一下-LMLPHP

[leetcode]25. Reverse Nodes in k-Group每k个节点反转一下-LMLPHP

[leetcode]25. Reverse Nodes in k-Group每k个节点反转一下-LMLPHP

[leetcode]25. Reverse Nodes in k-Group每k个节点反转一下-LMLPHP

code:

 /*
Time: O(n)
Space: O(1)
*/
class Solution {
public ListNode reverseKGroup(ListNode head, int k) {
if (head == null) return null;
ListNode dummy = new ListNode(-1);
ListNode pre = dummy;
dummy.next = head;
// to reverse each k-Group, considering pre as a "dummy" ahead
while (pre != null) {
pre = reverse(pre, k);
}
return dummy.next;
} public ListNode reverse(ListNode pre, int k) {
// to reverse each k-Group, considering last as a "null" boundary
ListNode last = pre;
for (int i = 0; i < k + 1; i++) {
last = last.next;
if (i != k && last == null) return null;
} // reverse
ListNode tail = pre.next;
ListNode cur = pre.next.next;
// remove cur to front, then update cur
while (cur != last) {
ListNode next = cur.next;
cur.next = pre.next;
pre.next = cur;
tail.next = next;
cur = next;
}
return tail;
}
}
04-26 17:21
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