最近在重新学习数据结构和算法的知识，数据结构和算法问题本身比较枯燥和乏味，而且比较难不容易掌握，但是属于程序员内功的一部分，学习起来容易上瘾。
1. 单链表定义
package algorithm.datastructors;/*** 单向链表* @author i324779*/
public class ListNode {private int data;private ListNode next;public ListNode(int data) {this.data = data;}public int getData() {return data;}public void setData(int data) {this.data = data;}public ListNode getNext() {return next;}public void setNext(ListNode next) {this.next = next;}
}
2, 单链表的一些常见操作
package algorithm.datastructors;import java.util.Stack;/*** 单链表的一些操作** @author i324779*/
public class ListNodeOperation {/*** 找到链表的中间节点* 使用两个指针。让第一个指针的移动速度是另一个的两倍。* 当第一个指针到达表尾时，另一个指针则指向中间节点。* Note：如果链表节点数为奇数，则第（n/2）个节点为中间节点。** @param head* @return*/ListNode findMiddle(ListNode head) {ListNode ptr1x, ptr2x;ptr1x = head;ptr2x = head;int i = 0;// 不断循环，直至达到表尾（next后继指针为null，就表示达到最后一个节点）while (ptr1x.getNext() != null) {if (i == 0) {ptr1x = ptr1x.getNext();i = 1;} else {// 两个指针都后移ptr1x = ptr1x.getNext();ptr2x = ptr2x.getNext();i = 0;}}return ptr2x;}/*** 从表尾开始输出链表。** @param head*/void printListFromEnd(ListNode head) {if (head == null) {return;}printListFromEnd(head.getNext());System.out.println(head.getData());}/*** 检查链表的长度是奇数还是偶数。* 使用一个在链表中每次向后移动两个节点的指针。* 最后，如果指针值为null，那么聊表长度为偶数；* 否则指针指向表尾节点，链表长度为奇数。** @param head* @return*/public int isLinkedListLengthEven(ListNode head) {while (head != null && head.getNext() != null) {head = head.getNext().getNext();}if (head == null) {return 0;} else {return 1;}}/*** 合并两个有序链表为一个新的有序链表。** @param node1* @param node2* @return*/public ListNode mergeList(ListNode node1, ListNode node2) {if (node1 == null) {return node2;}if (node2 == null) {return node1;}ListNode result;if (node1.getData() <= node2.getData()) {result = node1;result.setNext(mergeList(node1.getNext(), node2));} else {result = node2;result.setNext(mergeList(node2.getNext(), node1));}return result;}/*** 逆置单向链表。** @param head 表头* @return 逆置后的单向链表。*/ListNode reverseList(ListNode head) {ListNode temp = null;ListNode nextNode;while (head != null) {nextNode = head.getNext();head.setNext(temp);temp = head;head = nextNode;}return temp;}/*** 找到链表的倒数第n个结点* 使用一次链表扫描解决问题。* 使用两个指针pNthNode和pTemp。* 首先，两个指针都指向链表的头结点。仅当pTemp（沿着链表）进行了n次移动后，* pNthNode才开始移动。然后两个指针同时移动直至tTemp到达表尾。* 这时pNthNode指针所指的结点就是所求的结点，也就是链表的倒数第n个结点。** @param head* @param nthNode* @return*/ListNode nthNodeFromEnd(ListNode head, int nthNode) {ListNode pTemp = head;ListNode pNnthNode = null;for (int count = 1; count < nthNode; count++) {if (pTemp != null) {pTemp = pTemp.getNext();}}while (pTemp != null) {if (pNnthNode == null) {pNnthNode = head;} else {pNnthNode = pNnthNode.getNext();}pTemp = pTemp.getNext();}return pNnthNode;}/*** 给定两个有序单链表的头指针head1 和head2， 打印两个链表的公共部分。** @param head1 有序链表1* @param head2 有序链表2*/public void printCommonPart(ListNode head1, ListNode head2) {if (head1 == null || head2 == null) {System.out.println("Node is null");return;}System.out.println("Common Part: ");while (head1 != null && head2 != null) {if (head1.getData() < head2.getData()) {head1 = head1.getNext();} else if (head1.getData() > head2.getData()) {head2 = head2.getNext();} else {System.out.println(head1.getData());head1 = head1.getNext();head2 = head2.getNext();}}}/*** 判断是否有环** @param head 链表头结点* @return 是否有环*/public static boolean isCycle(ListNode head) {ListNode p1 = head;ListNode p2 = head;while (p2 != null && p2.getNext() != null) {p1 = p1.getNext();p2 = p2.getNext().getNext();if (p1 == p2) {return true;}}return false;}/*** 逐对逆置链表** @param head* @return*/private void reversePairRecursive(ListNode head) {if (head == null || head.getNext() == null) {return;}ListNode cur = head.getNext();ListNode pre = head;ListNode next;while (cur != null && cur.getNext() != null) {next = cur.getNext().getNext();pre.setNext(cur.getNext());cur.getNext().setNext(cur);cur.setNext(next);pre = cur;cur = next;}}/*** 判断一个链表是否为回文结构** @param head 单链表的头结点* @return true：回文结构；false：不是回文结构*/public boolean isPalindrome(ListNode head) {Stack<ListNode> stack = new Stack<>();ListNode cur = head;while (cur != null) {stack.push(cur);cur = cur.getNext();}while (head != null) {if (head.getData() != stack.pop().getData()) {return false;}head = head.getNext();}return true;}public static void main(String[] args) {ListNodeOperation operation = new ListNodeOperation();int i = 1;ListNode head = new ListNode();head.setNext(null);ListNode tmp;ListNode cur = head;for (; i < 8; i++) {tmp = new ListNode();tmp.setData(i);tmp.setNext(null);cur.setNext(tmp);cur = tmp;}System.out.println("顺序输出：");for (cur = head.getNext(); cur != null; cur = cur.getNext()) {System.out.print(cur.getData() + " ");}operation.reversePairRecursive(head);System.out.println("\n逆序输出：");for (cur = head.getNext(); cur != null; cur = cur.getNext()) {System.out.print(cur.getData() + " ");}}
}
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