Reverse Nodes in k-Group
描述
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.
You may not alter the values in the nodes, only nodes itself may be changed.
Only constant memory is allowed.
For 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
题解
package algorithms
import "github.com/ljun20160606/leetcode/algorithms"
//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.
//
//You may not alter the values in the nodes, only nodes itself may be changed.
//
//Only constant memory is allowed.
//
//For 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
/**
* Definition for singly-linked list.
* type ListNode struct {
* Val int
* Next *ListNode
* }
*/
func reverseKGroup(head *algorithms.ListNode, k int) *algorithms.ListNode {
start := head
for i := 0; i < k; i++ {
if head == nil {
return start
}
head = head.Next
}
newHead := reverseGroup(start, head)
start.Next = reverseKGroup(head, k)
return newHead
}
func reverseGroup(start, end *algorithms.ListNode) *algorithms.ListNode {
var newHead, tempHead *algorithms.ListNode
for start != end {
tempHead = start.Next
start.Next = newHead
newHead = start
start = tempHead
}
return newHead
}