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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