剑指 Offer 24. 反转链表-简单

难度：简单

题目描述：
反转一个单链表。

示例：

输入: 1->2->3->4->5->NULL
输出: 5->4->3->2->1->NULL

题目解析：
递归：

var reverseList = function(head) {
    if(!head || !head.next) return head
    var next = head.next
    // 递归反转
    var reverseHead = reverseList(next)
    // 变更指针
    next.next = head
    head.next = null
    return reverseHead
};

循环：

var reverseList = function(head) {
    if(!head || !head.next) return head
    var prev = null, curr = head
    while(curr) {
        // 用于临时存储 curr 后继节点
        var next = curr.next
        // 反转 curr 的后继指针
        curr.next = prev
        // 变更prev、curr 
        // 待反转节点指向下一个节点 
        prev = curr
        curr = next
    }
    head = prev
    return head
};
var reverseList = function(head) {
      if (head === null || head.next === null) return head;
      let reverse = null;
      let pre = null;
      while (head) {
        pre = head;
        head = head.next;
        pre.next = reverse;
        reverse = pre;
      }
      return reverse;
};

尾递归

var reverseList = function(head) {
    if(!head || !head.next) return head
    head = reverse(null, head)
    return head
};

var reverse = function(prev, curr) {
    if(!curr) return prev
    var next = curr.next
    curr.next = prev
    return reverse(curr, next)
};