链表001反转单链表

输入 1->2->3->4->null输出 5->4->3->2->1->null

思路

思路 把指针指向前一个元素&＃xff0c;后边的不断向前移动&＃xff0c;并改变方向

pre 就是cur的最前边的那位&＃xff08;pre&＃61;cur&＃xff09;
cur就是当前元素后面链表元素&＃xff08;cur&＃61;cur.Next&＃xff09;
cur.Next肯定是pre&＃xff08;curl.Next&＃61;pre&＃xff09;原因是cur赋值给了pre

type ListNode struct {Val intNext *ListNode
}
/**************************************************************************/
//反转链表递归实现
func reversListTwo(head *ListNode)*ListNode{if head &＃61;&＃61; nil || head.Next &＃61;&＃61; nil{return head}pre :&＃61; reversListTwo(head.Next)head.Next.Next &＃61; headhead.Next &＃61; nilreturn pre
}
//反转链表one
func reversList(head *ListNode)*ListNode{cur :&＃61; headvar pre *ListNode &＃61; nilfor cur !&＃61; nil{pre ,cur ,cur.Next &＃61; cur,cur.Next,pre//关键点}return pre
}func main(){head :&＃61; new(ListNode)head.Val &＃61; 1ln2 :&＃61; new(ListNode)ln2.Val &＃61; 2ln3 :&＃61; new(ListNode)ln3.Val &＃61; 3ln4 :&＃61; new(ListNode)ln4.Val &＃61; 4ln5 :&＃61; new(ListNode)ln5.Val &＃61; 5head.Next &＃61; ln2ln2.Next &＃61; ln3ln3.Next &＃61; ln4ln4.Next &＃61; ln5pre :&＃61; reversList(head)fmt.Printf("%#v\n",pre)fmt.Printf("%#v\n",pre.Next)fmt.Printf("%#v\n",pre.Next.Next)fmt.Printf("%#v\n",pre.Next.Next.Next)fmt.Printf("%#v\n",pre.Next.Next.Next.Next)fmt.Printf("%#v\n",pre.Next.Next.Next.Next.Next)}

验证

&main.ListNode{Val:5, Next:(*main.ListNode)(0xc000104210)}
&main.ListNode{Val:4, Next:(*main.ListNode)(0xc000104200)}
&main.ListNode{Val:3, Next:(*main.ListNode)(0xc0001041f0)}
&main.ListNode{Val:2, Next:(*main.ListNode)(0xc0001041e0)}
&main.ListNode{Val:1, Next:(*main.ListNode)(nil)}
(*main.ListNode)(nil)