Merge pull request #143 from quanxing/master

添加带头节点翻转和不带头节点翻转以及回文串的实现
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wangzheng0822 2018-11-12 10:44:14 +08:00 committed by GitHub
commit d1b6fea156
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@ -30,6 +30,9 @@ public class SinglyLinkedList {
return p;
}
//无头结点
//表头部插入
//这种操作将于输入的顺序相反逆序
public void insertToHead(int value) {
Node newNode = new Node(value, null);
insertToHead(newNode);
@ -44,6 +47,24 @@ public class SinglyLinkedList {
}
}
//顺序插入
//链表尾部插入
public void insertTail(int value){
Node newNode = new Node(value, null);
//空链表可以插入新节点作为head也可以不操作
if (head == null){
head = newNode;
}else{
Node q = head;
while(q.next != null){
q = q.next;
}
newNode.next = q.next;
q.next = newNode;
}
}
public void insertAfter(Node p, int value) {
Node newNode = new Node(value, null);
insertAfter(p, newNode);
@ -146,6 +167,123 @@ public class SinglyLinkedList {
System.out.println();
}
//判断true or false
public boolean TFResult(Node left, Node right){
Node l = left;
Node r = right;
System.out.println("left_:"+l.data);
System.out.println("right_:"+r.data);
while(l != null && r != null){
if (l.data == r.data){
l = l.next;
r = r.next;
continue;
}else{
break;
}
}
System.out.println("什么结果");
if (l==null && r==null){
System.out.println("什么结果");
return true;
}else{
return false;
}
}
// 判断是否为回文
public boolean palindrome(){
if (head == null){
return false;
}else{
System.out.println("开始执行找到中间节点");
Node p = head;
Node q = head;
if (p.next == null){
System.out.println("只有一个元素");
return true;
}
while( q.next != null && q.next.next != null){
p = p.next;
q = q.next.next;
}
System.out.println("中间节点" + p.data);
System.out.println("开始执行奇数节点的回文判断");
Node leftLink = null;
Node rightLink = null;
if(q.next == null){
// p 一定为整个链表的中点且节点数目为奇数
leftLink = inverseLinkList(p);
System.out.println("左边第一个节点"+leftLink.data);
System.out.println("右边第一个节点"+p.next.data);
rightLink = p;
}else{
//p q 均为中点
leftLink = inverseLinkList(p);
rightLink = q;
}
return TFResult(leftLink, rightLink);
}
}
//带结点的链表翻转
public Node inverseLinkList_head(Node p){
// Head 为新建的一个头结点
Node Head = new Node(9999,null);
// p 为原来整个链表的头结点,现在Head指向 整个链表
Head.next = p;
/*
带头结点的链表翻转等价于
从第二个元素开始重新头插法建立链表
*/
Node Cur = p.next;
p.next = null;
Node next = null;
while(Cur != null){
next = Cur.next;
Cur.next = Head.next;
Head.next = Cur;
System.out.println("first " + Head.data);
Cur = next;
}
// 返回左半部分的中点之前的那个节点
// 从此处开始同步像两边比较
return Head;
}
//无头结点的链表翻转
public Node inverseLinkList(Node p){
Node pre = null;
Node r = head;
System.out.println("z---" + r.data);
Node next= null;
while(r !=p){
next = r.next;
r.next = pre;
pre = r;
r = next;
}
r.next = pre;
// 返回左半部分的中点之前的那个节点
// 从此处开始同步像两边比较
return r;
}
public static Node createNode(int value) {
return new Node(value, null);
}
@ -163,5 +301,34 @@ public class SinglyLinkedList {
return data;
}
}
public static void main(String[]args){
SinglyLinkedList link = new SinglyLinkedList();
System.out.println("hello");
//int data[] = {1};
//int data[] = {1,2,3,1};
int data[] = {1,2,4,5};
//int data[] = {1,2,2,1};
//int data[] = {1,2,5,2,1};
for(int i =0; i < data.length; i++){
//link.insertToHead(data[i]);
link.insertTail(data[i]);
}
link.printAll();
Node p = link.inverseLinkList_head(link.head);
while(p != null){
System.out.println("aa"+p.data);
p = p.next;
}
//System.out.println("打印原始:");
//link.printAll();
//if (link.palindrome()){
// System.out.println("回文");
//}else{
// System.out.println("不是回文");
//}
}
}