剑指 Offer 06. 从尾到头打印链表
输入一个链表的头节点，从尾到头反过来返回每个节点的值（用数组返回）。
示例 1：
输入：head = [1,3,2]
输出：[2,3,1]
限制：
0 <= 链表长度 <= 10000
//递归
func reversePrint(head *ListNode) []int {
if head == nil {
return nil
}
return append(reversePrint(head.Next), head.Val)
}
//栈
func reversePrint(head *ListNode) []int {
if head == nil {
return nil
}
var pre *ListNode
for head != nil {
temp := head.Next
head.Next = pre
pre = head
head = temp
}
var res []int
for pre != nil {
res = append(res, pre.Val)
pre = pre.Next
}
return res
}
/**
* Definition for singly-linked list.
* type ListNode struct {
* Val int
* Next *ListNode
* }
*/
func reverseBookList(head *ListNode) []int {
if head == nil {return nil}
res := []int{}
for head != nil {
res = append(res, head.Val)
head = head.Next
}
for i, j := 0, len(res)-1; i<j; {
res[i], res[j] = res[j], res[i]
i++
j--
}
return res
}
剑指 Offer 24. 反转链表
定义一个函数，输入一个链表的头节点，反转该链表并输出反转后链表的头节点。
示例:
输入: 1->2->3->4->5->NULL
输出: 5->4->3->2->1->NULL
限制：
0 <= 节点个数 <= 5000
//栈
func reverseList(head *ListNode) *ListNode {
if head == nil || head.Next == nil {
return head
}
var newhead *ListNode
for head != nil {
temp := head.Next
head.Next = newhead
newhead = head
head = temp
}
return newhead
}
//递归
func reverseList(head *ListNode) *ListNode {
if head == nil || head.Next == nil {
return head
}
newHead := reverseList(head.Next)
head.Next.Next = head
head.Next = nil
return newHead
}
