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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 进阶: 你可以迭代或递归地反转链表。你能否用两种方法解决这道题?
来源:力扣(LeetCode) 链接:https://leetcode-cn.com/problems/reverse-linked-list 著作权归领扣网络所有。商业转载请联系官方授权,非商业转载请注明出处。
记录一个 next 表示下一个节点, cur 表示当前节点,prev 表示上一个节点, 在循环中不断的把 cur.next 赋值为 prev,然后 cur 前进为刚刚保存的 next 节点,直到 cur 为 null。
/** * Definition for singly-linked list. * function ListNode(val) { * this.val = val; * this.next = null; * } */ /** * @param {ListNode} head * @return {ListNode} */ let reverseList = function (head) { let prev = null let cur = head while (cur) { let next = cur.next cur.next = prev prev = cur cur = next } return prev }
The text was updated successfully, but these errors were encountered:
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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
进阶:
你可以迭代或递归地反转链表。你能否用两种方法解决这道题?
来源:力扣(LeetCode)
链接:https://leetcode-cn.com/problems/reverse-linked-list
著作权归领扣网络所有。商业转载请联系官方授权,非商业转载请注明出处。
思路
记录一个 next 表示下一个节点, cur 表示当前节点,prev 表示上一个节点, 在循环中不断的把 cur.next 赋值为 prev,然后 cur 前进为刚刚保存的 next 节点,直到 cur 为 null。
The text was updated successfully, but these errors were encountered: