剑指 Offer 06. 从尾到头打印链表(递归、逆置链表、头部动态插入)

题目
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

解法一：递归（递归本来就是一种栈，所以算做一类）

func reversePrint(head *ListNode) []int {if head == nil {return []int{}}if head.Next == nil {return []int{head.Val}}return append(reversePrint(head.Next), head.Val)
}

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解法二：逆置链表

func reversePrint(head *ListNode) []int {if head == nil {return []int{}}// 逆置链表fistNode, mindNode := head, head.NextfistNode.Next = nilfor {if mindNode == nil {break}endNode := mindNode.NextmindNode.Next = fistNodefistNode = mindNodemindNode = endNode}vals := make([]int,0)for {if fistNode == nil {return vals}vals = append(vals, fistNode.Val)fistNode = fistNode.Next}
}

在这里插入图片描述
解法三：头部动态插入（模拟切片动态扩容）

func reversePrint(head *ListNode) []int {if head == nil {return []int{}}vals := make([]int, 4)endIndex := len(vals)for {endIndex--if head == nil {break}vals[endIndex] = head.Valif endIndex == 0 {vals, endIndex = dilatation(vals)}head = head.Next}return vals[endIndex+1:]
}func dilatation(arr []int) ([]int, int) {lenArr := len(arr)restfulSize := lenArr + lenArr/2restful := make([]int, restfulSize)endIndex := len(restful)arrIndex := lenArr - 1for {if arrIndex < 0 {break}endIndex--restful[endIndex] = arr[arrIndex]arrIndex--}return restful, endIndex
}