package com.coding.basic;
import java.util.Stack;
public class LinkedList implements List {
private Node head;
private int size = 0;
public void add(Object o) {
Node newNode = new Node(o);
if (null == head) {
head = newNode;
size++;
return;
}
Node currNode = head;
while (null != currNode.next) {
currNode = currNode.next;
}
currNode.next = newNode;
size++;
}
public void add(int index, Object o) {
rangeCheck(index);
if (index == size) {
add(o);
return;
}
if (index == 0) {
Node newNode = new Node(o);
newNode.next = head;
head = newNode;
size++;
return;
}
Node newNode = new Node(o);
Node currNode = head;
for (int i = 0; i < index - 1; i++) {
currNode = currNode.next;
}
newNode.next = currNode.next;
currNode.next = newNode;
size++;
}
public Object get(int index) {
rangeCheck(index);
Node node = head;
for (int i = 0; i < index; i++) {
node = node.next;
}
return node;
}
private void rangeCheck(int index) {
if (index < 0 || index > size) {
throw new IndexOutOfBoundsException("Index: " + index + ", Size: " + size);
}
}
public Object remove(int index) {
rangeCheck(index);
if (index == 0) {
return this.removeFirst();
}
if (index == size - 1) {
return this.removeLast();
}
Node currNode = head;
for (int i = 0; i < index - 1; i++) {
currNode = currNode.next;
}
Node removeNode = currNode.next;
currNode.next = removeNode.next;
// removeNode = null;
size--;
return removeNode;
}
public void remove(Object obj) {
if (head == null) {
throw new NullPointerException();
}
// 如果要删除的结点是第一个,则把下一个结点赋值给第一个结点
if (head.data.equals(obj)) {
head = head.next;
size--;
} else {
Node pre = head; // 上一节点
Node cur = head.next; // 当前结点
while (cur != null) {
if (cur.data.equals(obj)) {
pre.next = cur.next;
size--;
}
pre = pre.next;
cur = cur.next;
}
}
}
public int size() {
return size;
}
public void addFirst(Object o) {
if (null == head) {
head = new Node(o);
size++;
return;
}
Node newNode = new Node(o);
newNode.next = head;
head = newNode;
size++;
}
public void addLast(Object o) {
this.add(o);
}
public Object removeFirst() {
if (null == head) {
return null;
}
Node currNode = head;
head = currNode.next;
size--;
return head;
}
public Object removeLast() {
if (null == head) {
return null;
}
if (null == head.next) {
Node currNode = head;
head = null;
size--;
return currNode;
}
Node currNode = head;
while (null != currNode.next) {
currNode = currNode.next;
}
currNode = null;
size--;
return null;
}
@Override
public String toString() {
StringBuffer sb = new StringBuffer();
sb.append("[");
Node node = head;
while (node != null) {
sb.append(node.data);
if (node.next != null) {
sb.append(",");
}
node = node.next;
}
sb.append("]");
return sb.toString();
}
public Iterator iterator() {
return new MyIterator();
}
private static class Node {
Object data;
Node next;
Node(Object data) {
this.data = data;
}
@Override
public String toString() {
return this.data.toString();
}
}
private class MyIterator implements Iterator {
int cursor = 0;
Node curNode;
@Override
public boolean hasNext() {
return cursor != size;
}
@Override
public Object next() {
if (curNode == null) {
curNode = head;
} else {
curNode = curNode.next;
}
cursor++;
return curNode;
}
}
/**
* 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() {
if (head == null || head.next == null) {
return;
}
java.util.Stack<Node> stack = new Stack<>();
Node curNode = head;
while (curNode != null) {
stack.push(curNode);
Node nextNode = curNode.next;
curNode.next = null; // 断开指向下一个元素的指针
curNode = nextNode;
}
head = stack.pop();
curNode = head;
while (!stack.isEmpty()) {
Node nextNode = stack.pop();
curNode.next = nextNode;
curNode = nextNode;
}
}
/**
* 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 num = size / 2;
for (int i = 0; i < num; i++) {
removeFirst();
}
}
/**
* 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) {
throw new IndexOutOfBoundsException();
}
int len = size - i >= length ? length : size - i;
int k = 0;
while (k < len) {
remove(i);
k++;
}
}
/**
* 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) {
int[] arr = new int[list.size()];
for (int i = 0; i < list.size(); i++) {
arr[i] = Integer.parseInt(get(Integer.parseInt(list.get(i).toString())).toString());
}
return arr;
}
/**
* 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) {
for (int i = 0; i < list.size(); i++) {
this.remove(list.get(i).toString());
}
}
/**
* 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() {
if (head == null || head.next == null) {
throw new RuntimeException("LinkedList is empty!");
}
Node pre = head;
Node cur = head;
while (cur.next != null) {
cur = cur.next;
Object data = pre.data;
while (cur.data == data) {
if (cur.next == null) {
pre.next = null;
break;
}
pre.next = cur.next;
size--;
cur = cur.next;
if (cur == null) {
break;
}
}
pre = pre.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 (head == null) {
return;
}
Node node = head;
int start = -1;
int end = -1;
int i = 0;
while (node != null) {
if ((start == -1) && (int) node.data <= min) {
start = i;
}
if ((int) node.data >= max) {
end = i;
break;
}
node = node.next;
i++;
}
if (start == -1) {
start = 0;
}
if (end == -1) {
end = size;
}
this.remove(start, end - start);
}
/**
* 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 null;
}
LinkedList result = new LinkedList();
int i1 = 0;
int i2 = 0;
while (i1 < this.size && i2 < list.size()) {
int value1 = Integer.valueOf(this.get(i1).toString());
int value2 = Integer.valueOf(list.get(i2).toString());
if (value1 == value2) {
result.add(value1);
i1++;
i2++;
} else if (value1 < value2) {
i1++;
} else {
i2++;
}
}
return result;
}
}