反转链表

206. 反转链表

leetcode 链接

思路：

1.定义一个 pre 节点，一个 current 节点指向 head

2.使用 current 节点循环遍历链表，每次循环将 先将 current.next 节点缓存起来为 temp， 然后将 current.next 节点指向 pre， 然后将 pre 指向当前节点 current，最后将 current 指向缓存的 temp 节点。

typescript 解法

/**
 * Definition for singly-linked list.
 * class ListNode {
 *     val: number
 *     next: ListNode | null
 *     constructor(val?: number, next?: ListNode | null) {
 *         this.val = (val===undefined ? 0 : val)
 *         this.next = (next===undefined ? null : next)
 *     }
 * }
 */

function reverseList(head: ListNode | null): ListNode | null {
  let pre: ListNode = null;
  let current: ListNode = head;
  while (current != null) {
    const temp: ListNode | null = current.next;
    current.next = pre;
    pre = current;
    current = temp;
  }
  return pre;
}

go 解法

/**
 * Definition for singly-linked list.
 * type ListNode struct {
 *     Val int
 *     Next *ListNode
 * }
 */
func reverseList(head *ListNode) *ListNode {
    var pre *ListNode
    current := head
    for current != nil {
        temp := current.Next
        current.Next = pre
        pre = current
        current = temp
    }
    return pre
}