剑指offer-06-翻转链表

一、翻转链表，输出链表

public ListNode reverseList(ListNode head) { ListNode pre = null; ListNode cur = head; while (cur != null){ ListNode next = cur.next; cur.next = pre; pre = cur; cur = next; } return pre; }

二、翻转链表，输出数组 输入一个链表的头节点，从尾到头反过来返回每个节点的值（用数组返回）。

输入：head = [1,3,2] 输出：[2,3,1]

方法1：

先计算链表的长度，存放至数组中

public static int[] reverseList(ListNode head) { //记录链表长度 int length = 0; ListNode curNode = head; while (curNode != null){ length ++; curNode = curNode.next; } int[] res = new int[length]; for (int i = res.length - 1; i >= 0; i--) { res[i] = head.val; head = head.next; } return res; }

方法2

创建栈，将元素存放至栈中，先进后出存放至数组中

public int[] reversePrint(ListNode head) { Stack<ListNode> stack = new Stack<ListNode>(); while (head != null){ stack.push(head); head = head.next; } int[] arr = new int[stack.size()]; for (int i = 0; i < arr.length; i++) { arr[i] = stack.pop().val; } return arr; }