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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