链表的基本操作
定义
class Node{
Item item;//表示任意的数据类型
Node next;//指向下一个节点的引用
}
从链表头部插入元素
Node oldFirst = first;//oldFirst指向原来的头结点
first = new Node();
first.item = item;
first.next = oldFirst;
从链表头部删除元素
first = first.next; //first指向first.next即可
从链表尾部插入元素
Node oldLast = last;
last = new Node();
last.item = item;
oldLast.next = last;
从中间插入和删除结点
Node first = new Node();
Node third = new Node();
first.next = third;
//在first和second之间插入second
Node second = new Node();
first.next = second;
second.next = third;
//删除second结点(删除结点必须知道前一个结点)
first.next = third;
栈和对列的实现
栈的实现
使用数组实现栈
public class ArrayStack<Item> implements Iterable<Item>{
//初始化数组的大小
private Item[] arr = (Item[])new Object[5];
private int N == 0;
public int size(){
return N;
}
public boolean isEmpty(){
return N == 0 ? true : false;
}
public void push(Item item){
if(N == arr.length){
resize(2*N);
}
arr[N++] = item;
}
public Item pop(){
Item item = arr[--N];
arr[N] = null;
if(N > 0 && N == arr.length/4){
resize(2*N);
}
return item;
}
//动态调整数组大小
public void resize(int capacity){
Item[] temp = (Item[])new Object[capacity];
for(int i = 0; i < N; i++){
temp[i] = arr[i];
}
arr = temp;
}
//实现迭代器功能，可以使用foreach语句
public Iterator<Item> iterator(){
return new ArrayReverseIterator();
}
private class ArrayReverseIterator implements Iterator<Item>{
private int index = N;
public boolean hasNext(){
return N > 0 ? true : false;
}
public Item next(){
return arr[--N];
}
}
}
使用链表实现栈
public class NodeStack<Item> implements Iterable<Item> {
private class Node{
Item item;
Node next;
}
private Node first;
private int N = 0;
public void push(Item item){
Node oldfirst = first;
first = new Node();
first.item = item;
first.next = oldfirst;
N++;
}
public Item pop(){
Item item = first.item;
first = first.next;
N--;
return item;
}
public boolean isEmpty(){
return N == 0;
}
public int size(){
return N;
}
@Override
public Iterator<Item> iterator() {
return new ListIterator();
}
private class ListIterator implements Iterator<Item>{
private Node current = first;
@Override
public boolean hasNext() {
return current != null;
}
@Override
public Item next() {
Item item = current.item;
current = current.next;
return item;
}
}
}
队列的实现
public class Queue<Item> implements Iterable<Item>{
private class Node{
Item item;
Node next;
}
Node first;
Node last;
private int N = 0;
public boolean isEmpty(){
return N == 0;
}
public int size(){
return N;
}
public void enqueue(Item item){
Node oldLast = last;
last = new Node();
last.item = item;
if(isEmpty()){
first = last;
}else{
oldLast.next = last;
}
N++;
}
public Item dequeue(){
Item item = first.item;
first = first.next;
if(isEmpty()){
last = null;
}
N--;
return item;
}
public Iterator<Item> iterator(){
return new QueueIterator();
}
private class QueueIterator implements Iterator<Item>{
private Node current = first;
public boolean hasNext(){
return current != null;
}
public Item next(){
Item item = current.item;
current = current.next;
return item;
}
}
}
链表练习题目
1.有一个整数val，如何在节点值有序的环形链表中插入一个节点值为val的节点，并且保证这个环形单链表依然有序.给定链表的信息，及元素的值A及对应的nxt指向的元素编号同时给定val，请构造出这个环形链表，并插入该值。
class ListNode{
int val;
ListNode next;
public ListNode(int val){
this.val = val;
}
}
public ListNode insert(int[] A, int[] next, int val){
if(A == null || A.length == 0){
ListNode node = new ListNode(val);
return node;
}
ListNode head = new ListNode(A[0]);
ListNode point = head;
ListNode current = null;
//通过数组构造链表
for(int i = 0; i < next.length - 1; i++){
current = new ListNode(A[next[i]]);
ponit.next = current;
point = current;
}
if(val < head.val){
ListNode node = new ListNode(val);
node.next = head;
head = node;
return head;
}
if(val > current.val){
ListNode node = new ListNode(val);
current.next = node;
return head;
}
ListNode p1 = head;
ListNode p2 = head.next;
while(p1 != null && p2 != null){
if(p1.val <= val && val <= p2.val){
ListNode node = new ListNode(val);
p1.next = node;
node.next = p2;
return head;
}else{
p1 = p1.next;
p2 = p2.next;
}
}
return head;
}
2.实现一个算法，删除单向链表中间的某个结点，假定你只能访问该结点。给定带删除的节点，请执行删除操作，若该节点为尾节点，返回false，否则返回true.
public boolean remove(ListNode pNode){
if(pNode.next == null){
return false;
}
pNode.val = pNode.next.val;
pNode.next = pNode.next.next;
return true;
}
3.对于一个链表，我们需要用一个特定阈值完成对它的分化，使得小于等于这个值的结点移到前面，大于该值的结点在后面，同时保证两类结点内部的位置关系不变。给定一个链表的头结点head，同时给定阈值val，请返回一个链表，使小于等于它的结点在前，大于等于它的在后，保证结点值不重复。
public ListNode listDivide(ListNode head,int val){
public ListNode head_1 = null;
public ListNode tail_1 = null;
public ListNode head_2 = null;
public ListNode tail_2 = null;
public ListNode next = null;
while(head != null){
next = head.next;
head.next = null;
if(head.val <= val){
ListNode oldTail_1 = tail_1;
tail_1 = head;
if(head_1 == null){
head_1 = tail_1;
}else{
oldtail_1.next = tail_1;
}
}else if(head.val > val){
ListNode oldTail_2 = tail_2;
tail_2 = head;
if(head_2 == null){
head_2 = tail_2;
}else{
oldTail_2.next = tail_2;
}
}
head = next;
}
if(tail_1 != null){
tail_1.next = head_2;
}
return head_1 != null ? head_1 : head_2;
}
4.有一个单链表，请设计一个算法，使得每K个节点之间逆序，如果最后不够K个节点一组，则不调整最后几个节点。例如链表1->2->3->4->5->6->7->8->null，K=3这个例子。调整后为，3->2->1->6->5->4->7->8->null。因为K==3，所以每三个节点之间逆序，但其中的7，8不调整，因为只有两个节点不够一组。
public ListNode inverse(ListNode head,int k){
if(head == null || head.next == null || k < 2){
return head;
}
ListNode pre = null;
ListNode current = head;
ListNode start = null;
ListNode next = null;
int count = 1;
while(current != null){
next = current.next;
if(count == 3){
start = pre == null ? head : pre.next;
head = pre == null ? current : head;
reverse(pre,start,current,next);
pre = start;
count = 0;
}
count++;
current = next;
}
return head;
}
//先逆序k个节点
public void reverse(ListNode left,ListNode start,ListNode end,ListNode right){
ListNode current = start.next;
ListNode point = start;
ListNode next = null;
while(current != right){
next = current.next;
current.next = point;
point = current;
current = next;
}
if(left != null){
left.next = end;
}
start.next = right;
}
5.如何判断一个单链表是否有环？有环的话返回进入环的第一个节点的值，无环的话返回-1。
public int chkLoop(ListNode head, int adjust) {
if(head == null){
return -1;
}
boolean isCircle = false;
ListNode fast = head;
ListNode slow = head;
while(slow != null && fast != null && fast.next != null){
slow = slow.next;
fast = fast.next.next;
if(slow == fast){
isCircle = true;
break;
}
}
if(!isCircle){
return -1;
}
fast = head;
while(slow != fast){
slow = slow.next;
fast = fast.next;
}
return slow.val;
}
6.现在有两个无环单链表，若两个链表的长度分别为m和n，请设计一个时间复杂度为O(n + m)， 额外空间复杂度为O(1)的算法，判断这两个链表是否相交。
public boolean chkIntersect(ListNode headA, ListNode headB) {
if(headA == null || headB == null){
return false;
}
boolean isIntersect = false;
int lenA = 0;
ListNode p1 = headA;
while(p1 != null){
p1 = p1.next;
lenA++;
}
int lenB = 0;
ListNode p2 = headB;
while(p2 != null){
p2 = p2.next;
lenB++;
}
if(lenA >= lenB){
isIntersect = isCrossing(headA,headB,lenA - lenB);
}else{
isIntersect = isCrossing(headB,headA,lenB - lenA);
}
return isIntersect;
}
public boolean isCrossing(ListNode headA, ListNode headB, int num){
ListNode curA = headA;
boolean isCross = false;
while(num != 0){
curA = curA.next;
num--;
}
ListNode curB = headB;
while(curA != null && curB != null){
if(curA == curB){
isCross = true;
break;
}
curA = curA.next;
curB = curB.next;
}
return isCross;
}
7.如何判断两个有环单链表是否相交？相交的话返回第一个相交的节点，不想交的话返回空。
public boolean chkInter(ListNode head1, ListNode head2, int adjust0, int adjust1) {
if(head1 == null || head2 == null){
return false;
}
ListNode p1 = insert(head1);
ListNode p2 = insert(head2);
//入环节点相交
if(p1 == p2){
return true;
}
//环中相交
ListNode start1 = p1.next;
while(start1 != p1){
if(start1 == p2){
return true;
}
start1 = start1.next;
}
return false;
}
//入环节点
public ListNode insert(ListNode head){
ListNode slow = head;
ListNode fast = head;
while(slow != null && fast != null){
slow = slow.next;
fast = fast.next.next;
if(slow == fast){
break;
}
}
fast = head;
while(slow != fast){
slow = slow.next;
fast = fast.next;
}
return slow;
}
