package com.aaront.exercise.basic;
public class LinkedList implements List {
private Node head = new Node(null);
private int size = 0;
public void add(Object o) {
Node newNode = new Node(o);
Node first = head.next;
Node second = head;
while (first != null) {
second = first;
first = first.next;
}
second.next = newNode;
size++;
}
public void add(int index, Object o) {
if (index < 0 || index > size) throw new IndexOutOfBoundsException("索引超出范围");
Node first = head;
int i = 0;
while (i < index) {
first = first.next;
i++;
}
Node node = new Node(o);
node.next = first.next;
first.next = node;
size++;
}
public Object get(int index) {
if (index < 0 || index >= size) throw new IndexOutOfBoundsException("索引超出范围");
Node first = head.next;
int i = 0;
while (i < index) {
first = first.next;
i++;
}
return first.data;
}
public Object remove(int index) {
if (index < 0 || index >= size) throw new IndexOutOfBoundsException("索引超出范围");
Node first = head;
int i = 0;
while (i < index) {
first = first.next;
i++;
}
Node element = first.next;
first.next = first.next.next;
size--;
return element.data;
}
public int size() {
return size;
}
public void addFirst(Object o) {
add(0, o);
}
public void addLast(Object o) {
add(size, o);
}
public Object removeFirst() {
return remove(0);
}
public Object removeLast() {
return remove(size - 1);
}
public Iterator iterator() {
return new LinkedListIterator(this);
}
public Object[] toArray() {
Object[] objects = new Object[size];
Node first = head.next;
int pos = 0;
while (first != null) {
objects[pos++] = first.data;
first = first.next;
}
return objects;
}
private static class LinkedListIterator implements Iterator {
private int pos = 0;
private LinkedList linkedList;
private LinkedListIterator(LinkedList linkList) {
this.linkedList = linkList;
}
@Override
public boolean hasNext() {
return pos < linkedList.size();
}
@Override
public Object next() {
return linkedList.get(pos++);
}
@Override
public void remove() {
linkedList.remove(pos - 1);
pos--;
}
}
private static class Node {
private Object data;
private Node next;
private Node(Object data) {
this.data = data;
}
}
/**
* 把该链表逆置
* 例如链表为 3->7->10 , 逆置后变为 10->7->3
*/
public void reverse() {
Node newHead = new Node(null);
Node node = head.next;
while (node != null) {
Node temp = node.next;
node.next = newHead.next;
newHead.next = node;
node = temp;
}
head = newHead;
}
/**
* 删除一个单链表的前半部分
* 例如：list = 2->5->7->8 , 删除以后的值为 7->8
* 如果list = 2->5->7->8->10 ,删除以后的值为7,8,10
*/
public void removeFirstHalf() {
int removeLen = size / 2;
if (size <= 1) {
head.next = null;
size = 0;
return;
}
for (int i = 0; i < removeLen; i++) {
Node point = head.next;
head.next = point.next;
point.next = null;
size--;
}
}
/**
* 从第i个元素开始， 删除length 个元素 ， 注意i从0开始
*
* @param i
* @param length
*/
public void remove(int i, int length) {
if (i < 0 || i >= size || length < 1) throw new IndexOutOfBoundsException("索引超出范围");
int endIndex = Math.min(i + length, size);
Node preNode = head;
Node endNode = head.next;
for (int index = 0; index < endIndex; index++) {
if (index < i) {
preNode = endNode;
}
endNode = endNode.next;
}
preNode.next = endNode;
size = size - (endIndex - i);
}
/**
* 假定当前链表和list均包含已升序排列的整数
* 从当前链表中取出那些list所指定的元素
* 例如当前链表 = 11->101->201->301->401->501->601->701
* listB = 1->3->4->6
* 返回的结果应该是[101,301,401,601]
*
* @param list
*/
public int[] getElements(LinkedList list) {
if (list == null) return new int[0];
Iterator iterator = list.iterator();
int[] result = new int[list.size()];
int index = 0;
while (iterator.hasNext()) {
Integer next = (Integer) iterator.next();
result[index] = (Integer) this.get(next);
index++;
}
return result;
}
/**
* 已知链表中的元素以值递增有序排列，并以单链表作存储结构。
* 从当前链表中中删除在list中出现的元素
*
* @param list
*/
public void subtract(LinkedList list) {
if (list == null) return;
Iterator iterator = this.iterator();
while (iterator.hasNext()) {
Object element = iterator.next();
if (contain(element, list)) {
iterator.remove();
}
}
}
private boolean contain(Object element, LinkedList list) {
Iterator iterator = list.iterator();
while (iterator.hasNext()) {
Object next = iterator.next();
if (next == element) return true;
}
return false;
}
/**
* 已知当前链表中的元素以值递增有序排列，并以单链表作存储结构。
* 删除表中所有值相同的多余元素（使得操作后的线性表中所有元素的值均不相同）
*/
public void removeDuplicateValues() {
Node pre = head;
Node cur = head.next;
while (cur != null) {
if (pre.data == cur.data) {
Node node = cur.next;
while (node != null && node.data == pre.data) {
node = node.next;
size--;
}
pre.next = node;
cur = node;
size--;
} else {
pre = cur;
cur = cur.next;
}
}
}
/**
* 已知链表中的元素以值递增有序排列，并以单链表作存储结构。
* 试写一高效的算法，删除表中所有值大于min且小于max的元素（若表中存在这样的元素）
*
* @param min
* @param max
*/
public void removeRange(int min, int max) {
if (min >= max) return;
Node cur = head;
Node start = null;
Node end = null;
Integer startIndex = null;
Integer endIndex = size - 1;
Integer index = -1;
while (cur.next != null) {
if (start == null && (Integer) cur.next.data > min) {
start = cur;
startIndex = index;
}
if ((Integer) cur.next.data >= max) {
end = cur.next;
endIndex = index;
break;
}
cur = cur.next;
index++;
}
if (start != null) {
start.next = end;
size = size - (endIndex - startIndex);
}
}
/**
* 假设当前链表和参数list指定的链表均以元素依值递增有序排列（同一表中的元素值各不相同）
* 现要求生成新链表C，其元素为当前链表和list中元素的交集，且表C中的元素有依值递增有序排列
*
* @param list
*/
public LinkedList intersection(LinkedList list) {
if (list == null) return this;
LinkedList intersection = new LinkedList();
Node node1 = this.head.next;
Node node2 = list.head.next;
while (node1 != null && node2 != null) {
if((Integer)node1.data < (Integer)node2.data) {
node1 = node1.next;
} else if((Integer)node1.data > (Integer)node2.data) {
node2 = node2.next;
} else {
intersection.add(node1.data);
node1 = node1.next;
node2 = node2.next;
}
}
return intersection;
}
}