java 实现单链表逆转详解及实例代码

java 实现单链表逆转详解

实例代码：

class Node {
  Node next;
  String name;
  public Node(String name) {
    this.name = name;
  }

  /**
   * 打印结点
   */
  public void show() {
    Node temp = this;
    do {
      System.out.print(temp + "->");
      temp = temp.next;
    }while(temp != null);
    System.out.println();
  }

  /**
   * 递归实现单链表反转,注意：单链表过长，会出现*Error
   * @param n
   * @return
   */
  public static Node recursionReverse(Node n) {
    long start = System.currentTimeMillis();
    if(n == null || n.next == null) {
      return n;
    }
    Node reverseNode = recursionReverse(n.next);

    n.next.next = n;
    n.next = null;
    System.out.println("递归逆置耗时:" + (System.currentTimeMillis() - start) + "ms...");
    return reverseNode;
  }

  /**
   * 循环实现单链表反转
   * @param n
   * @return
   */
  public static Node loopReverse(Node n) {
    long start = System.currentTimeMillis();
    if(n == null || n.next == null) {
      return n;
    }

    Node pre = n;
    Node cur = n.next;
    Node next = null;
    while(cur != null) {
      next = cur.next;
      cur.next = pre;
      pre = cur;
      cur = next;
    }
    n.next = null;
    n = pre;
    System.out.println("循环逆置耗时:" + (System.currentTimeMillis() - start) + "ms...");
    return pre;
  }

  @Override
  public String toString() {
    return name;
  }

　　public static void main(String[] args) {

    int len = 10;
    Node[] nodes = new Node[len];
    for(int i = 0; i < len; i++) {
      nodes[i] = new Node(i + "");
    }
    for(int i = 0; i < len - 1; i++) {
      nodes[i].next = nodes[i+1];
    }
    /* try {
      Thread.sleep(120000);
    } catch (InterruptedException e) {
      e.printStackTrace();
    }*/
    Node r1 = Node.loopReverse(nodes[0]);
    r1.show();
    Node r = Node.recursionReverse(r1);
    r.show();

  }
}

总结

对于递归和循环，推荐使用循环实现，递归在单链表过大时，会出现StatckOverflowError,递归涉及到方法的调用，在性能上也弱于循环的实现

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原文链接：http://asflex.iteye.com/blog/2084962