java 单链表反转

链接：https://www.jianshu.com/p/d201e6a00e3f

.准备链表

准备一个由DataNode组成的单向链表，DataNode如下：

public class DataNode {

    private int data; private DataNode next; public int getData() { return data; } public void setData(int data) { this.data = data; } public DataNode getNext() { return next; } public void setNext(DataNode next) { this.next = next; } public DataNode(int data) { this.data = data; } } 

构造链表

public class DataChain {
    
    private DataNode head; public DataChain(int size) { DataNode head = new DataNode(0); DataNode cur = head; for (int i = 1; i < size; i++) { DataNode tmp = new DataNode(i); cur.setNext(tmp); cur = tmp; } this.head = head; } public DataNode getHead() { return head; } public void setHead(DataNode head) { this.head = head; } public static void printChain(DataNode head) { StringBuilder sb = new StringBuilder(); DataNode cur = head; sb.append(cur.getData()); while (null != cur.getNext()) { sb.append(" -> "); sb.append(cur.getNext().getData()); cur = cur.getNext(); } System.out.println(sb.toString()); } public static void main(String... strings) { DataChain chain = new DataChain(10); printChain(chain.getHead()); } } 

运行main方法，即构造了一个包含10个node节点的单链表。

#运行结果
0 -> 1 -> 2 -> 3 -> 4 -> 5 -> 6 -> 7 -> 8 -> 9

2.通过递归实现单链表反转

考虑到代码的简洁性，首先考虑的是通过递归实现。

    /**
     * 递归实现 当栈深度大于12000 则会出现StakOverflowError
     * 
     * @param head
     * @return
     */
    public static DataNode reverse1(DataNode head) { if (null == head || null == head.getNext()) return head; DataNode revHead = reverse1(head.getNext()); head.getNext().setNext(head); head.setNext(null); return revHead; } 

以上即是递归实现的源码，但是需要考虑的问题是递归都在java栈中进行，需要考虑jdk支持的栈的深度。在jdk1.8.0_91版本中，当上述链表长度大于12000则会出现StackOverFlowError错误。说明对于该版本jdk栈的深度不能大于12000。

3.通过遍历实现

最通用的实现方式就是遍历。

/**
     * 遍历实现 通用实现方法
     * 
     * @param head
     * @return
     */
    public static DataNode reverse2(DataNode head) { if (null == head || null == head.getNext()) return head; DataNode pre = head; DataNode cur = head.getNext(); while (null != cur.getNext()) { DataNode tmp = cur.getNext(); cur.setNext(pre); pre = cur; cur = tmp; } cur.setNext(pre); head.setNext(null); return cur; } 

4.借助stack实现

考虑到stack具有先进后出这一特性，因此可以借助于stack数据结构来实现单向链表的反转。

/**
     * 方法3 利用其他数据结构 stack 
     * @param head
     * @return
     */
    public static DataNode reverse3(DataNode head) { Stack<DataNode> stack = new Stack<DataNode>(); for (DataNode node = head; null != node; node = node.getNext()) { stack.add(node); } DataNode reHead = stack.pop(); DataNode cur = reHead; while(!stack.isEmpty()){ cur.setNext(stack.pop()); cur = cur.getNext(); cur.setNext(null); } return reHead; } 

上述实现方法在于操作简单，对于算法并不精通的同学可以尝试。缺点在于需要通过其他数据结构实现，效率会降低，至于效率会降低到什么程度，后面举例说明。

5.三种实现方式效率分析

    public static void main(String... strings) { int size = 10; DataChain chain1 = new DataChain(size); printChain(chain1.getHead()); long reverse1_start = System.currentTimeMillis(); DataNode reNode1 = reverse1(chain1.getHead()); long reverse1_cost = System.currentTimeMillis() - reverse1_start; printChain(reNode1); System.out.println("reverse1 cost time is ["+reverse1_cost+"]ms"); DataChain chain2 = new DataChain(size); printChain(chain2.getHead()); long reverse2_start = System.currentTimeMillis(); DataNode reNode2 = reverse2(chain2.getHead()); long reverse2_cost = System.currentTimeMillis() - reverse2_start; printChain(reNode2); System.out.println("reverse2 cost time is ["+reverse2_cost+"]ms"); DataChain chain3 = new DataChain(size); printChain(chain3.getHead()); long reverse3_start = System.currentTimeMillis(); DataNode reNode3 = reverse3(chain3.getHead()); long reverse3_cost = System.currentTimeMillis() - reverse3_start; printChain(reNode3); System.out.println("reverse3 cost time is ["+reverse3_cost+"]ms"); } 

执行结果：

0 -> 1 -> 2 -> 3 -> 4 -> 5 -> 6 -> 7 -> 8 -> 9 9 -> 8 -> 7 -> 6 -> 5 -> 4 -> 3 -> 2 -> 1 -> 0 reverse1 cost time is [0]ms 0 -> 1 -> 2 -> 3 -> 4 -> 5 -> 6 -> 7 -> 8 -> 9 9 -> 8 -> 7 -> 6 -> 5 -> 4 -> 3 -> 2 -> 1 -> 0 reverse2 cost time is [0]ms 0 -> 1 -> 2 -> 3 -> 4 -> 5 -> 6 -> 7 -> 8 -> 9 9 -> 8 -> 7 -> 6 -> 5 -> 4 -> 3 -> 2 -> 1 -> 0 reverse3 cost time is [1]ms 

在上述代码基础上，去掉打印输出，将size改为10000，结果如下：

reverse1 cost time is [1]ms
reverse2 cost time is [0]ms
reverse3 cost time is [6]ms 

可以看出reverse2 明显优于其他两种实现方法。考虑到reverse1最多只支持12000，因此将size改为100000时，再观察reverse2和reverse3之间的执行结果：

reverse2 cost time is [6]ms
reverse3 cost time is [25]ms

因此可以看出，最好的方法是采用遍历的方式进行反转。