0025. Reverse Nodes in K Group

25. Reverse Nodes in k-Group #

题目 #

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.

题目大意 #

按照每 K 个元素翻转的方式翻转链表。如果不满足 K 个元素的就不翻转。

解题思路 #

这一题是 problem 24 的加强版,problem 24 是两两相邻的元素,翻转链表。而 problem 25 要求的是 k 个相邻的元素,翻转链表,problem 相当于是 k = 2 的特殊情况。

代码 #


package leetcode

/**
 * Definition for singly-linked list.
 * type ListNode struct {
 *     Val int
 *     Next *ListNode
 * }
 */
func reverseKGroup(head *ListNode, k int) *ListNode {
	node := head
	for i := 0; i < k; i++ {
		if node == nil {
			return head
		}
		node = node.Next
	}
	newHead := reverse(head, node)
	head.Next = reverseKGroup(node, k)
	return newHead
}

func reverse(first *ListNode, last *ListNode) *ListNode {
	prev := last
	for first != last {
		tmp := first.Next
		first.Next = prev
		prev = first
		first = tmp
	}
	return prev
}



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