简体   繁体   English

查找单个链表的中间

[英]Finding middle of single linked list

I had modified my function to take care of the case for even number of nodes, but because of that my code does not work for an odd number of nodes. 我修改了函数以照顾偶数个节点的情况,但是由于这个原因,我的代码不适用于奇数个节点。 What is wrong? 怎么了?

public LinkedList findMiddleNode() {
    Node t1 = this.getHeadNode();
    Node t2 = this.getHeadNode();
    LinkedList l = new LinkedList();
    boolean even = false;
    while(t1.getNext() != null) {
        t1 = t1.getNext();
        if(t1.getNext()!= null && t1.getNext().getNext() != null) {
            t1 = t1.getNext();
            t2 = t2.getNext();              
        }
    }
    if(t1.getNext()!=null) 
        l.insertFirst(t2.getElement());
    else {
        l.insertFirst(t2.getElement());
        l.insertLast(t2.getNext().getElement());
    }
    return l;
}

@oldrinb: thnx for ur help, here is my updated code: @oldrinb:thnx为您提供帮助,这是我的更新代码:

public LinkedList findMiddleNode() {
    Node t1 = this.getHeadNode();
    Node t2 = this.getHeadNode();
    LinkedList l = new LinkedList();
    boolean even = false;
    while(t1.getNext() != null && t1.getNext().getNext() != null) {
            t1 = t1.getNext().getNext();
            t2 = t2.getNext();              
    }
    if(t1.getNext()==null) 
        l.insertFirst(t2.getElement());
    else {
        l.insertFirst(t2.getElement());
        l.insertLast(t2.getNext().getElement());
    }
    return l;
}
/* determine length of list */
int n = 0;
for (Node cur = getHeadNode(); cur != null; cur = cur.getNext()) {
  ++n;
}

/* go half-way */
int m = n / 2;
Node mid = getHeadNode();
while (m > 0) {
  mid = mid.getNext();
  --m;
}

/* if n is odd, both mid and mid.getNext() are the middle */
if ((n % 2) != 0) {
  Node mid2 = mid.getNext();
  ...
} else {
  ...
}

... or, alternatively, ...或者,

Node tortoise = getHeadNode();
Node hare = getHeadNode();

do {
  tortoise = tortoise.getNext();
} while (hare.getNext() != null && (hare = hare.getNext().getNext()) != null);

/* tortoise is the middle */

声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.

 
粤ICP备18138465号  © 2020-2024 STACKOOM.COM