数据结构之单链表的几个简单算法题

        单链表作为最基本的数据结构，在程序设计中有着非常重要的运用。最近自己闲下来，正在整理数据结构和算法的一些程序题，现将自己的代码贴出来与大家分享。如有不对之处，请大家指正。（好吧 ，这么简单的题目应该不会有错，都测试过了。况且说的好像很多人看我博客一样。好久没更，都长草了……就当做写给自己看的吧。:P）

package test;import java.util.Hashtable;class LinkNode{int value;LinkNode next;public LinkNode(int d){value = d;}public LinkNode(int d, LinkNode n){value = d;next = n;}/** * 打印链表 */public static void printLinkedList(LinkNode head){LinkNode p = head;while( null != p ){System.out.print("" + p.value + " ");p = p.next;}System.out.println();}}/** * LinkNodeOperation * 1.删除链表中重复的节点  * 2.找出单链表中倒数的第K个元素 * 3.实现单链表的反转 * 4.实现单链表逆向输出 * 5.寻找单链表中间节点 * 6.判断两条链表是否相交 */public class LinkNodeOperation {public static void main(String[] args){// 生成一条链表：LinkNode n4 = new LinkNode(19,null);LinkNode n3 = new LinkNode(20,n4);LinkNode n2 = new LinkNode(30,n3);LinkNode n1 = new LinkNode(40,n2);LinkNode head1 = new LinkNode(50,n1);// 生成一条值为0-20的链表,头结点为2：LinkNode head = new LinkNode(2);LinkNode cur = null;for( int i=1; i<19; i++){LinkNode temp = new LinkNode(i);if( i==1 ){ head.next = temp;}else{cur.next = temp;}cur = temp;}cur.next = n4;//打印原始链表：System.out.print("原始链表：");LinkNode.printLinkedList(head);//打印原始链表：System.out.print("待判断链表：");LinkNode.printLinkedList(head1);//判断两条链表是否相交：System.out.println("是否相交："+ LinkNodeOperation.isIntersect(head, head1));//寻找链表的中间节点：System.out.print("链表的中间节点：");LinkNodeOperation.findMiddleElem(head);//从尾到头输出：System.out.print("从尾到头输出单链表：");    LinkNodeOperation.printLinklistReversely(head);    System.out.println();        //打印倒数第K个节点：    LinkNodeOperation.findElem(head, 4);    System.out.println("----------------------------------------");    LinkNodeOperation.findElem2(head, 4);    System.out.println("----------------------------------------");        //打印去重后的链表：    System.out.print("去重后的链表：");    LinkNodeOperation.deleteDuplicate2(head);        //打印反转后的链表    LinkNode.printLinkedList(head);    System.out.print("反转后的链表：");    LinkNodeOperation.reverseElem(head);    }/** * 使用HashTable去重 */public static void deleteDuplicate(LinkNode head){Hashtable<Integer, Integer> hashTable = new Hashtable<Integer, Integer>();LinkNode tmp = head;LinkNode pre = null;while(tmp != null){if(hashTable.containsKey(tmp.value)){pre.next = tmp.next;}else{hashTable.put(tmp.value, 1);pre = tmp;}tmp = tmp.next;}}/** * 使用常规方法：双重循环遍历 */public static void deleteDuplicate2(LinkNode head){LinkNode p = head;while(p!=null){LinkNode q = p;while(q.next!=null){if(p.value == q.next.value){q.next = q.next.next;}else{q = q.next;}}p = p.next;}}/** * 找出倒数第K个元素 * 方法1两次遍历 */public static void findElem(LinkNode head,int k){int count = 1;int index = 0;LinkNode p = head;while(p.next!= null){count++;p = p.next;}if( k<1 || k>count ){return;}System.out.println("--------------------------------");System.out.println("链表长度为："+ count);while( head != null){index++;if(index == count-k+1){break;}head = head.next;}System.out.println("倒数第" + k + "个节点为：" + head.value);}/** * 找出倒数第K个元素 * 方法2 双指针 只需一次遍历 */public static void findElem2(LinkNode head, int k){if(k<1){return;}LinkNode p = head;LinkNode q = head;// 让P比Q先走K-1步for(int i=1; i<=k-1; i++){p = p.next;}while(p.next != null){p = p.next;q = q.next;}System.out.println("倒数第" + k +"个节点是：" + q.value );}    /**     * 链表反转     */public static void reverseElem(LinkNode head){LinkNode pre = head;LinkNode cur = head.next;LinkNode next= null;while(null != cur){next = cur.next;cur.next = pre;pre = cur;cur = next;}head.next = null;head = pre;LinkNode.printLinkedList(head);}/** * 从尾到头输出单链表 * 方法：递归 */public static void printLinklistReversely(LinkNode head){if(head != null){printLinklistReversely(head.next);System.out.print(""+head.value+" ");}}/** * 寻找单链表的中间节点 */public static void findMiddleElem(LinkNode head){LinkNode p = head;LinkNode q = head;while(p!=null && p.next!=null && p.next.next!=null ){p = p.next.next;q = q.next;}System.out.println(""+q.value);}/** * 判断两条链表是否相交 */public static boolean isIntersect(LinkNode h1, LinkNode h2){if (h1==null || h2==null) {return false;}LinkNode tail1 = h1;LinkNode tail2 = h2;while(tail1.next!=null){tail1 = tail1.next;}while(tail2.next!=null){tail2 = tail2.next;}return tail1==tail2;}}

运行结果：