LeetCode刷题之反转链表

• 反转链表是数据结构中较为简单的题目，其核心是依靠三个中间变量，进行链表的反转
• 用next记录下一个节点，用pre记录当前节点前一个节点
• 当前current节点不为空，则进行循环
• 过程为
  • 假设有A、B、C、D三个节点，此时进行循环，head.next不为空
  • next = head.next;head.next = pre;pre = head;head = next;
  • 假设为A节点，next和pre的初始值都为null;
  • next = head.next;A.next = null;pre = A;A = next;
• 这里模拟
  • 假设一个链表：A-> B -> C，此时反转为C->B->A
  • head.next = next head.next -> B
    • 然后next = A.next; next -> B next指向B
    • head.next = pre，此时head.next -> null(pre)
    • pre = A, pre -> A，pre指向A
    • head = next, head-> next，head节点指向next(B)
  • 第二次循环
    • next = head.next next -> C
    • head.next = pre B.next -> A
    • pre = head pre-> B
    • head = next head -> C
  • 第三次循环，此时next不为null
    • next = head.next; next -> C.next(null)
    • head.next = pre; c.next -> B
    • pre = head; pre -> C
    • head = next; head -> next
  • 再次判断head为null结束循环
  • 返回pre
• code

var reverseHead = function (head) {
 if (head === null || head.next === null) {
   return head
 }
 let next = null
 let pre = null
 while(head !== null) {
   next = head.next
   head.next = pre
   pre = head
   head = next
 }
 return pre
}