package link;
import list.ArrayList;
import list.List;
import java.util.*;
/**
* Created by gongxun on 2017/3/13.
*/
public class LinkedList<T> {
private Node<T> head;
private int size = 0;
public void add(T o) {
if (head == null) {
head = new Node<T>(null, o);
size++;
} else
addLast(o);
}
private Node<T> getLast() {
Node<T> last = head;
while (last.next != null) {
last = last.next;
}
return last;
}
private Node<T> getNodeIndex(int index) {
if (index > size - 1)
throw new IndexOutOfBoundsException("size : " + size + ", index : " + index);
Node target = head;
for (int i = 0; i < size; i++) {
if (i == index)
return target;
target = target.next;
}
return null;
}
public void add(int index, T o) {
Node<T> node = getNodeIndex(index - 1);
Node<T> nextNode = node.next;
node.next = new Node<T>(nextNode, o);
size++;
}
public T get(int index) {
Node<T> node = getNodeIndex(index);
return node.data;
}
public T remove(int index) {
Node<T> prev = getNodeIndex(index - 1);
Node<T> now = getNodeIndex(index);
prev.next = now.next;
size--;
return now.data;
}
public int size() {
return size;
}
public void addFirst(T o) {
if (head != null)
head = new Node<T>(null, o);
else {
Node newNode = new Node(head, o);
head = newNode;
}
size++;
}
public void addLast(T o) {
add(size, o);
}
public T removeFirst() {
Node<T> removeNode = head;
if (head != null)
head = head.next;
size--;
return removeNode == null ? null : removeNode.data;
}
public T removeLast() {
Node<T> last = getNodeIndex(size - 1);
Node<T> prev = getNodeIndex(size - 2);
prev.next = null;
size--;
return last.data;
}
public Iterator iterator() {
return null;
}
private int getIndex(Node<T> node) {
Node temp = head;
int index = 0;
while (temp != null) {
if (temp == node) {
return index;
}
}
return -1;
}
private int getIndexByData(T data) {
Node temp = head;
int index = 0;
while (temp != null) {
if ((data == null && temp.data == null) || temp.data.equals(data))
return index;
index++;
temp = temp.next;
}
return -1;
}
private static class Node<T> {
T data;
Node<T> next;
Node(Node next, T data) {
this.next = next;
this.data = data;
}
}
@Override
public String toString() {
StringBuilder sb = new StringBuilder();
Node temp = head;
while (temp != null) {
sb.append(temp.data).append("-->");
temp = temp.next;
}
return sb.toString().substring(0, sb.lastIndexOf("-->"));
}
/**
* 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 cur = null;
Node prev = null;
while (head != null) {
cur = head;
head = head.next;
cur.next = prev;
prev = cur;
}
head = cur;
}
/**
* 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() {
head = getNodeIndex(size / 2);
}
/**
* 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 (size <= (i + length) || i < 0)
throw new IndexOutOfBoundsException("size : " + size + ", i + length : " + (i + length));
Node<T> rightNode = getNodeIndex(i + length);
if (i == 0)
head = rightNode;
else {
Node<T> leftNode = getNodeIndex(i - 1);
leftNode.next = rightNode;
}
}
/**
* 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 Object[] getElements(LinkedList<Integer> list) {
Object[] result = new Object[list.size];
if (list != null) {
for (int i = 0; i < list.size; i++) {
result[i] = get(list.get(i));
}
}
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<T> list) {
if (list != null && list.size > 0) {
for (int i = 0; i < list.size; i++) {
int index = getIndexByData(list.get(i));
if (index != -1)
remove(index);
else
throw new RuntimeException("wrong element of removed list");
}
}
}
/**
* 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<T> temp = head;
List<T> list = new ArrayList<T>();
int index = 0;
while (temp != null) {
if (list.contains(temp.data))
remove(index--);
else
list.add(temp.data);
temp = temp.next;
index++;
}
}
/**
* 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) {
Integer first = (Integer) get(0);
Integer last = (Integer) getLast().data;
if (first > max || last < min)
return;
List<Integer> indexRange = new ArrayList<Integer>();
Node<Integer> temp = (Node<Integer>) head;
int index = 0;
while (temp != null) {
if (temp.data >= min && temp.data <= max) {
indexRange.add(index);
}
index++;
temp = temp.next;
}
remove(indexRange.get(0), indexRange.size());
}
/**
* 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<Integer> list) {
LinkedList<Integer> newList = new LinkedList<Integer>();
merge(newList, (Node<Integer>) this.head, list.head);
return newList;
}
private void merge(LinkedList<Integer> newList, Node<Integer> thisHead, Node<Integer> mergeHead) {
if (thisHead == null && mergeHead == null)
return;
if (thisHead == null) {
//无论是否包含,有元素的链表必须指向next
if (!newList.contains(mergeHead.data))
newList.add(mergeHead.data);
mergeHead = mergeHead.next;
merge(newList, null, mergeHead);
}
if (mergeHead == null) {
if (!newList.contains(thisHead.data))
newList.add(thisHead.data);
thisHead = thisHead.next;
merge(newList, thisHead, null);
}
//要再进行一次判断是因为递归到最底层return之后,返回上一层时某个链表已经为null了,但是上一层还是会将剩下的执行完
if (thisHead != null && mergeHead != null) {
if (thisHead.data < mergeHead.data && !newList.contains(thisHead.data)) {
newList.add(thisHead.data);
thisHead = thisHead.next;
merge(newList, thisHead, mergeHead);
} else if (!newList.contains(mergeHead.data)) {
newList.add(mergeHead.data);
mergeHead = mergeHead.next;
merge(newList, thisHead, mergeHead);
}
}
}
private boolean contains(Integer data) {
int index = this.getIndexByData((T) data);
return index != -1;
}
}