LeetCode 206 Reverse Linked List 反转链表
<pre><code class="language-js">// 递归处理, 利用函数作用域操作当前节点和当前节点的下一个节点const reverseList = head => {
if (head == null || head.next == null) return head
const end = reverseList(head.next)
head.next.next = head
head.next = null
return end// 返回最后一个节点，也就是反转后的头节点
};
const reverseList = head => {
let prev = null
let cur = head
while (cur) {
const next = cur.next
cur.next = prev
prev = cur
cur = next
}
return prev
}
</code></pre><br><br>
来源：https://www.cnblogs.com/ltfxy/p/16493970.html
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