242 lines
6.1 KiB
Java
242 lines
6.1 KiB
Java
今天咱们说一道非常简单但是很经典的面试题，思路很容易，但是里面细节挺多，所以我们还是需要注意。
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我们先来看一下题目描述
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#### [206. 反转链表](https://leetcode-cn.com/problems/reverse-linked-list/)
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反转一个单链表。
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**示例:**
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> 输入: 1->2->3->4->5->NULL
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> 输出: 5->4->3->2->1->NULL
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该题目我们刚开始刷题的同学可能会想到先保存到数组中，然后从后往前遍历数组，重新组成链表，这样做是可以 AC 的，但是我们机试时往往不允许我们修改节点的值，仅仅是修改节点的指向。所以我们应该用什么方法来解决呢？
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我们先来看动图，看看我们能不能理解。然后再对动图进行解析。
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原理很容易理解，我们首先将 low 指针指向空节点， pro 节点指向 head 节点，
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然后我们定义一个临时节点指向 pro 节点，
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此时我们就记住了 pro 节点的位置，然后 pro = pro.next.这样我们三个指针指向三个不同的节点。
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则我们将 temp 指针指向 low 节点，此时则完成了反转。
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反转之后我们继续反转下一节点，则 low = temp 即可。然后重复执行上诉操作直至最后，这样则完成了反转链表。
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我们下面看代码吧。
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我会对每个关键点进行注释，大家可以参考动图理解。
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**题目代码**
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Java Code:
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```java
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class Solution {
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public ListNode reverseList(ListNode head) {
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//特殊情况
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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 low = null;
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ListNode pro = head;
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while (pro != null) {
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//代表橙色指针
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ListNode temp = pro;
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//移动绿色指针
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pro = pro.next;
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//反转节点
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temp.next = low;
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//移动黄色指针
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low = temp;
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}
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return low;
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}
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}
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```
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JS Code:
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```javascript
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var reverseList = function(head) {
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//特殊情况
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if(!head || !head.next) {
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return head;
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}
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let low = null;
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let pro = head;
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while (pro) {
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//代表橙色指针
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let temp = pro;
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//移动绿色指针
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pro = pro.next;
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//反转节点
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temp.next = low;
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//移动黄色指针
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low = temp;
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}
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return low;
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};
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```
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C++ Code:
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```cpp
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class Solution {
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public:
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ListNode* reverseList(ListNode* head) {
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//特殊情况
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if (head == nullptr || head->next == nullptr) {
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return head;
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}
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ListNode * low = nullptr;
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ListNode * pro = head;
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while (pro != nullptr) {
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//代表橙色指针
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ListNode * temp = pro;
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//移动绿色指针
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pro = pro->next;
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//反转节点
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temp->next = low;
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//移动黄色指针
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low = temp;
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}
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return low;
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}
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};
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```
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Python Code:
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```py
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class Solution:
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def reverseList(self, head: ListNode) -> ListNode:
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//特殊情况
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if head is None or head.next is None:
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return head
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low = None
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pro = head
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while pro is not None:
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# 代表橙色指针
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temp = pro
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# 移动绿色指针
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pro = pro.next
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# 反转节点
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temp.next = low
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# 移动黄色指针
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low = temp
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return low
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```
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上面的迭代写法是不是搞懂啦，现在还有一种递归写法，不是特别容易理解，刚开始刷题的同学，可以只看迭代解法。
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**题目代码**
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Java Code:
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```java
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class Solution {
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public ListNode reverseList(ListNode head) {
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//结束条件
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if (head == null || head.next == null) {
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return head;
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}
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//保存最后一个节点
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ListNode pro = reverseList(head.next);
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//将节点进行反转。我们可以这样理解 4.next.next = 4;
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//4.next = 5；
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//则 5.next = 4 则实现了反转
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head.next.next = head;
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//防止循环
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head.next = null;
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return pro;
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}
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}
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```
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JS Code:
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```javascript
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var reverseList = function(head) {
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//结束条件
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if (!head || !head.next) {
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return head;
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}
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//保存最后一个节点
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let pro = reverseList(head.next);
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//将节点进行反转。我们可以这样理解 4.next.next = 4;
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//4.next = 5；
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//则 5.next = 4 则实现了反转
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head.next.next = head;
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//防止循环
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head.next = null;
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return pro;
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};
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```
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C++代码:
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```cpp
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class Solution {
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public:
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ListNode * reverseList(ListNode * head) {
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//结束条件
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if (head == nullptr || head->next == nullptr) {
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return head;
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}
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//保存最后一个节点
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ListNode * pro = reverseList(head->next);
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//将节点进行反转。我们可以这样理解 4->next->next = 4;
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//4->next = 5；
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//则 5->next = 4 则实现了反转
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head->next->next = head;
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//防止循环
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head->next = nullptr;
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return pro;
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}
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};
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```
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Python Code:
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```py
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class Solution:
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def reverseList(self, head: ListNode) -> ListNode:
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# 结束条件
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if head is None or head.next is None:
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return head
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# 保存最后一个节点
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pro = self.reverseList(head.next)
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# 将节点进行反转。我们可以这样理解 4->next->next = 4;
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# 4->next = 5；
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# 则 5->next = 4 则实现了反转
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head.next.next = head
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# 防止循环
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head.next = None
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return pro
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```
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<br/>
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> 贡献者[@jaredliw](https://github.com/jaredliw)注：
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>
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> 这里提供一个比较直观的递归写法供大家参考。
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>
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> ```py
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> class Solution:
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> def reverseList(self, head: ListNode, prev_nd: ListNode = None) -> ListNode:
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> # 结束条件
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> if head is None:
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> return prev_nd
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> # 记录下一个节点并反转
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> next_nd = head.next
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> head.next = prev_nd
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> # 给定下一组该反转的节点
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> return self.reverseList(next_nd, head)
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> ```
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