php实现单向链表反转

<?php


//思路：就地反转
//把当前链表的下一个节点插入到头结点的下一个节点中。

pCur是需要反转的节点。

1. prev连接下一次需要反转的节点    prev.next = pCur.next;
2. 反转节点pCur   　　                  pCur.next = dummy.next;
3. 纠正头结点dummy的指向            dummy.next = pCur;
4. pCur指向下一次要反转的节点      pCur = prev.next;

结束条件为pCur!=null

class Node{
    private $value;
    private $next;

    public function __construct($value=null){
        $this->value = $value;
    }
    public function getValue(){
        return $this->value;
    }
    public function setValue($value){
        $this->value = $value;
    }
    public function getNext(){
        return $this->next;
    }
    public function setNext($next){
        $this->next = $next;
    }
}
//遍历，将当前节点的下一个节点缓存后更改当前节点指针 
function reverse($head){
    if($head == null){
        return $head;
    }
    $pre = $head;//注意：对象的赋值
    $cur = $head->getNext();
    $next = null;
    while($cur != null){
        $next = $cur->getNext();
        $cur->setNext($pre);
        $pre = $cur;
        $cur = $next;
    }
    //将原链表的头节点的下一个节点置为null，再将反转后的头节点赋给head  
    $head->setNext(null);
    $head = $pre;
    return $head;
}
//递归，在反转当前节点之前先反转后续节点 
function reverse2($head){
    if (null == $head || null == $head->getNext()) {
        return $head;
    }
    $reversedHead = reverse2($head->getNext());
    $head->getNext()->setNext($head);
    $head->setNext(null);
    return $reversedHead;
}
function test(){
    $head = new Node(0);
    $tmp = null;
    $cur = null;
    // 构造一个长度为10的链表，保存头节点对象head   
    for($i=1;$i<10;$i++){
        $tmp = new Node($i);
        if($i == 1){
            $head->setNext($tmp);
        }else{
            $cur->setNext($tmp);
        }
        $cur = $tmp;
    }
    //print_r($head);exit;
    $tmpHead = $head;
    while($tmpHead != null){
        echo $tmpHead->getValue().' ';
        $tmpHead = $tmpHead->getNext();
    }
    echo "\n";
    //$head = reverse($head);
    $head = reverse2($head);
    while($head != null){
        echo $head->getValue().' ';
        $head = $head->getNext();
    }
}
test();
?>