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---
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链表相关
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---
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[06. 从头到尾打印链表](https://leetcode-cn.com/problems/cong-wei-dao-tou-da-yin-lian-biao-lcof/)
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```java
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class Solution {
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public int[s] reversePrint(ListNode head) {
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ListNode temp = head;
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int size = 0;
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while (temp != null) {
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temp = temp.next;
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size++;
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}
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int[] result = new int[size];
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int index = size - 1;
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while (head != null) {
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result[index--] = head.val;
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head = head.next;
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}
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return result;
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}
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}
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```
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```java
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class Solution {
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public int[] reversePrint(ListNode head) {
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Stack<ListNode> stack = new Stack<ListNode>();
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ListNode temp = head;
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while (temp != null) {
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stack.push(temp);
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temp = temp.next;
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}
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int size = stack.size();
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int[] print = new int[size];
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for (int i = 0; i < size; i++) {
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print[i] = stack.pop().val;
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}
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return print;
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}
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}
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```
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[22. 链表中倒数第 k 个节点](https://leetcode-cn.com/problems/lian-biao-zhong-dao-shu-di-kge-jie-dian-lcof/)
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```java
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class Solution {
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public ListNode getKthFromEnd(ListNode head, int k) {
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int length = 0;
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ListNode temp = head;
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while (temp != null) {
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temp = temp.next;
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length++;
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}
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for (int i = 0; i < length - k; i++) {
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head = head.next;
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}
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return head;
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}
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}
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```
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```java
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class Solution {
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public ListNode getKthFromEnd(ListNode head, int k) {
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ListNode h0 = head;
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for (int i = 0; i < k; i++) {
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h0 = h0.next;
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}
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while (h0 != null) {
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h0 = h0.next;
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head = head.next;
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}
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return head;
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}
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}
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```
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[24. 反转链表](https://leetcode-cn.com/problems/fan-zhuan-lian-biao-lcof/)
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```java
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class Solution {
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public ListNode reverseList(ListNode head) {
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if (head == null || head.next == null) {
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return head;
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}
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ListNode h1 = head;
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ListNode h2 = head.next;
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ListNode h3 = null;
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h1.next = null;
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while (h2 != null) {
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h3 = h2.next;
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h2.next = h1;
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h1 = h2;
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h2 = h3;
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}
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return h1;
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}
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}
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```
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```java
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class Solution {
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public ListNode reverseList(ListNode head) {
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// 递归终止条件是当前为空,或者下一个节点为空
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if (head == null || head.next == null) {
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return head;
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}
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// 这里的 h1 就是最后一个节点
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ListNode h1 = reverseList(head.next);
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// 如果链表是 1->2->3->4->5,那么此时的 cur 就是 5
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// 而 head 是4,head的 下一个是 5,下下一个是空
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// 所以 head.next.next 就是 5->4
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head.next.next = head;
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// 防止链表循环,需要将 head.next 设置为空
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head.next = null;
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// 每层递归函数都返回 h1,也就是最后一个节点
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return h1;
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}
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}
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```
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