Task1.1 数组
1.1.1实现一个支持动态扩容的数组
public class EnsureCapacityArray {
private static final Object[] EMPTY_ELEMENTDATA = {};
private static final Object[] DEFAULTCAPACITY_EMPTY_ELEMENTDATA = {};
transient Object[] elementData;
//传入固定值的情况
public EnsureCapacityArray(int initialCapacity) {
if (initialCapacity > 0) {
this.elementData = new Object[initialCapacity];
} else if (initialCapacity == 0) {
this.elementData = EMPTY_ELEMENTDATA;
} else {
throw new IllegalArgumentException("Illegal Capacity: "+
initialCapacity);
}
}
//没有传入数组大小的情况
public EnsureCapacityArray() {
this.elementData = DEFAULTCAPACITY_EMPTY_ELEMENTDATA;
}
private static final int MAX_ARRAY_SIZE = Integer.MAX_VALUE - 8;
/**
* 扩容
* @param minCapacity
*/
public void grow(int minCapacity) {
int oldCapacity = elementData.length;
int newCapacity = oldCapacity + (oldCapacity >> 1);
if (newCapacity - minCapacity < 0)
newCapacity = minCapacity;
if (newCapacity - MAX_ARRAY_SIZE > 0)
newCapacity = hugeCapacity(minCapacity);
// minCapacity is usually close to size, so this is a win:
elementData = Arrays.copyOf(elementData, newCapacity);
}
//数组容量最大的情况
private static int hugeCapacity(int minCapacity) {
if (minCapacity < 0) // overflow
throw new OutOfMemoryError();
return (minCapacity > MAX_ARRAY_SIZE) ?
Integer.MAX_VALUE :
MAX_ARRAY_SIZE;
}
}1.1.2 实现一个大小固定的有序数组，支持动态增删改操作
这里理解，大小固定的有序数组，那也就是说数组大小和内部的元素个数完全相同。那么增的时候，需要扩容，删的时候，需要减容。然后数组有序，也就是说，改的时候需要重新排序。
public class FixedArray{
private static final int DEFAULT_SIZE = 10;
private static final Object[] EMPTY_ELEMENTDATA = {};
private static final Object[] DEFAULT_ELEMENTDATA = new Object[DEFAULT_SIZE];
transient Object[] elementData;
private int size;
public FixedArray(){
this.elementData=DEFAULT_ELEMENTDATA;
}
public FixedArray(int initialCapacity){
if (initialCapacity > 0) {
this.elementData = new Object[initialCapacity];
} else if (initialCapacity == 0) {
this.elementData = EMPTY_ELEMENTDATA;
} else {
throw new IllegalArgumentException("Illegal Capacity: "+
initialCapacity);
}
}
//保证增和删时数组长度和元素长度一致
private void ensureCapacityInternal(int minCapacity ){
if (minCapacity - elementData.length > 0)
grow(minCapacity);
}
private static final int MAX_ARRAY_SIZE = Integer.MAX_VALUE - 8;
public void grow(int minCapacity) {
int oldCapacity = elementData.length;
int newCapacity = oldCapacity + (oldCapacity >> 1);
if (newCapacity - minCapacity < 0)
newCapacity = minCapacity;
if (newCapacity - MAX_ARRAY_SIZE > 0)
newCapacity = hugeCapacity(minCapacity);
// minCapacity is usually close to size, so this is a win:
elementData = Arrays.copyOf(elementData, newCapacity);
}
//数组容量最大的情况
private static int hugeCapacity(int minCapacity) {
if (minCapacity < 0) // overflow
throw new OutOfMemoryError();
return (minCapacity > MAX_ARRAY_SIZE) ?
Integer.MAX_VALUE :
MAX_ARRAY_SIZE;
}
//去掉删除之后，重新排序的，空的数组
public void trimToSize() {
if (size < elementData.length) {
elementData = (size == 0)
? EMPTY_ELEMENTDATA
: Arrays.copyOf(elementData, size);
}
}
//判断数组长度
public int size(){
return size;
}
//判断数组是否为空
public boolean isEmpty() {
return size == 0;
}
//排序
@SuppressWarnings("unchecked")
private <E> void sort() {
Arrays.sort((E[]) elementData, 0, size);
}
//添加
public <E> boolean add(E e) {
ensureCapacityInternal(size + 1);
elementData[size++] = e;
sort();
return true;
}
//删除
public <E> boolean remove(int index) {
if (index >= size)
throw new IndexOutOfBoundsException();
System.arraycopy(elementData, index+1, elementData, index,
size-1);
size--;
trimToSize();
return true;
}
//改
public <E> boolean set(int index,Object e) {
elementData[index] = e;
sort();
return true;
}
//查
public Object get(int index) {
return elementData[index];
}
}1.1.3 合并两个有序的数组
public static int[] mergeArrays(int[]a ,int[]b){
int alen=a.length;
int blen=b.length;
int aindex=0;
int bindex=0;
int [] re=new int[alen+blen];
for(int i=0;i<alen+blen;i++){
if(aindex<alen && bindex<blen){
re[i]=a[aindex] > b[bindex] ? b[bindex++] :a[aindex++];
continue;
}
if(aindex==alen && bindex<blen )
re[i]=b[bindex++];
if(bindex==blen &&aindex<alen)
re[i]=a[aindex++];
}
return re;
}1.1.4学习哈希表思想，并完成leetcode上的两数之和(1)及Happy Number(202)！
哈希表
//两数之和
public int[] twoSum(int[] nums, int target) {
Map<Integer, Integer> map = new HashMap<>();
for (int i = 0; i < nums.length; i++) {
int complement = target - nums[i];
if (map.containsKey(complement)) {
return new int[] { map.get(complement), i };
}
map.put(nums[i], i);
}
throw new IllegalArgumentException("No two sum solution");
}
//Happy Number(202)
class Solution {
public boolean isHappy(int n) {
if(n==1) return true;
while(n!=1 && n!=4){
int a=0;
int ans=0;
while(n>0){
a=n%10;
ans+=a*a;
n /=10;
}
n=ans;
}
if(n==1) return true;
else return false;
}
}1.1.5 练习
//Three Sum
public List<List<Integer>> threeSum(int[] num) {
Arrays.sort(num);
List<List<Integer>> res = new LinkedList<>();
for (int i = 0; i < num.length-2; i++) {
if (i == 0 || (i > 0 && num[i] != num[i-1])) {
int lo = i+1, hi = num.length-1, sum = 0 - num[i];
while (lo < hi) {
if (num[lo] + num[hi] == sum) {
res.add(Arrays.asList(num[i], num[lo], num[hi]));
while (lo < hi && num[lo] == num[lo+1]) lo++;
while (lo < hi && num[hi] == num[hi-1]) hi--;
lo++; hi--;
} else if (num[lo] + num[hi] < sum) lo++;
else hi--;
}
}
}
return res;
}
//Majority Element
public int majorityElement(int[] nums) {
int res=0;
int cnt=0;
for(int num:nums){
if(cnt==0){
res=num;
++cnt;
}
else if(num == res) ++cnt;
else --cnt;
}
return res;
}
//Missing Positive
public int firstMissingPositive(int[] nums) {
if(nums == null&&nums.length==0)
return 1;
int i;
for(i=0;i<nums.length;i++){
if(nums[i] != i+1){
while(0<nums[i]&& nums[i] <= nums.length && nums[nums[i]-1] != nums[i]){
int temp = nums[i];
nums[i] = nums[nums[i]-1];
nums[temp-1] = temp;
}
}
}
for(i=0;i<nums.length;i++) {
if(nums[i]!=i+1) {
return i+1;
}
}
return i+1;
}TASK1.2 链表
1.2.1 实现单链表、循环链表、双向链表，支持增删操作
//单链表
class singleNode{
public int data;
public singleNode next;
public singleNode head =null;
public singleNode(int data){
this.data=data;
}
public void addNode(singleNode node){
singleNode newNode = new singleNode(data);
if(head == null){
head = newNode;
return;
}
singleNode temp = head;
while(temp.next != null){
temp = temp.next;
}
temp.next = newNode;
}
public boolean removeNode(int index){
if(index<1 || index>length()){
return false;
}
if(index == 1){//删除头结点
head = head.next;
return true;
}
singleNode preNode = head;
singleNode curNode = preNode.next;
int i = 1;
while(curNode != null){
if(i==index){//寻找到待删除结点
preNode.next = curNode.next;//待删除结点的前结点指向待删除结点的后结点
return true;
}
//当先结点和前结点同时向后移
preNode = preNode.next;
curNode = curNode.next;
i++;
}
return true;
}
public int length(){
int length = 0;
singleNode curNode = head;
while(curNode != null){
length++;
curNode = curNode.next;
}
return length;
}
}//双向链表
//单向循环链表类
public class CycleLinkList implements List {
Node head; //头指针
Node current;//当前结点对象
int size;//结点个数
//初始化一个空链表
public CycleLinkList()
{
//初始化头结点，让头指针指向头结点。并且让当前结点对象等于头结点。
this.head = current = new Node(null);
this.size =0;//单向链表，初始长度为零。
this.head.next = this.head;
}
//定位函数，实现当前操作对象的前一个结点，也就是让当前结点对象定位到要操作结点的前一个结点。
//比如我们要在a2这个节点之前进行插入操作，那就先要把当前节点对象定位到a1这个节点，然后修改a1节点的指针域
public void index(int index) throws Exception
{
if(index <-1 || index > size -1)
{
throw new Exception("参数错误！");
}
//说明在头结点之后操作。
if(index==-1) //因为第一个数据元素结点的下标是0，那么头结点的下标自然就是-1了。
return;
current = head.next;
int j=0;//循环变量
while(current != head&&j<index)
{
current = current.next;
j++;
}
}
@Override
public void delete(int index) throws Exception {
// TODO Auto-generated method stub
//判断链表是否为空
if(isEmpty())
{
throw new Exception("链表为空，无法删除！");
}
if(index <0 ||index >size)
{
throw new Exception("参数错误！");
}
index(index-1);//定位到要操作结点的前一个结点对象。
current.setNext(current.next.next);
size--;
}
@Override
public Object get(int index) throws Exception {
// TODO Auto-generated method stub
if(index <-1 || index >size-1)
{
throw new Exception("参数非法！");
}
index(index);
return current.getElement();
}
@Override
public void insert(int index, Object obj) throws Exception {
// TODO Auto-generated method stub
if(index <0 ||index >size)
{
throw new Exception("参数错误！");
}
index(index-1);//定位到要操作结点的前一个结点对象。
current.setNext(new Node(obj,current.next));
size++;
}
@Override
public boolean isEmpty() {
// TODO Auto-generated method stub
return size==0;
}
@Override
public int size() {
// TODO Auto-generated method stub
return this.size;
}
}//双向循环链表
//单向链表类
public class DoubleCycleLinkList implements List {
Node head; //头指针
Node current;//当前结点对象
int size;//结点个数
//初始化一个空链表
public DoubleCycleLinkList() {
//初始化头结点，让头指针指向头结点。并且让当前结点对象等于头结点。
this.head = current = new Node(null);
this.size = 0;//单向链表，初始长度为零。
this.head.next = head;
this.head.prior = head;
}
//定位函数，实现当前操作对象的前一个结点，也就是让当前结点对象定位到要操作结点的前一个结点。
public void index(int index) throws Exception {
if (index < -1 || index > size - 1) {
throw new Exception("参数错误！");
}
//说明在头结点之后操作。
if (index == -1)
return;
current = head.next;
int j = 0;//循环变量
while (current != head && j < index) {
current = current.next;
j++;
}
}
@Override
public void delete(int index) throws Exception {
// TODO Auto-generated method stub
//判断链表是否为空
if (isEmpty()) {
throw new Exception("链表为空，无法删除！");
}
if (index < 0 || index > size) {
throw new Exception("参数错误！");
}
index(index - 1);//定位到要操作结点的前一个结点对象。
current.setNext(current.next.next);
current.next.setPrior(current);
size--;
}
@Override
public Object get(int index) throws Exception {
// TODO Auto-generated method stub
if (index < -1 || index > size - 1) {
throw new Exception("参数非法！");
}
index(index);
return current.getElement();
}
@Override
public void insert(int index, Object obj) throws Exception {
// TODO Auto-generated method stub
if (index < 0 || index > size) {
throw new Exception("参数错误！");
}
index(index - 1);//定位到要操作结点的前一个结点对象。
current.setNext(new Node(obj, current.next));
current.next.setPrior(current);
current.next.next.setPrior(current.next);
size++;
}
@Override
public boolean isEmpty() {
// TODO Auto-generated method stub
return size == 0;
}
@Override
public int size() {
// TODO Auto-generated method stub
return this.size;
}
}1.2.2 实现单链表反转
public ListNode reverseList(ListNode head) {
if (head == null || head.next == null) return head;
ListNode p = reverseList(head.next);
head.next.next = head;
head.next = null;
return p;
}1.2.3 实现两个有序的链表合并为一个有序链表
ListNode* mergeTwoLists(ListNode* l1, ListNode* l2) {
if(l1 == NULL) return l2;
if(l2 == NULL) return l1;
if(l1->val < l2->val) {
l1->next = mergeTwoLists(l1->next, l2);
return l1;
} else {
l2->next = mergeTwoLists(l2->next, l1);
return l2;
}
}1.2.4 实现求链表的中间结点
public ListNode middleNode(ListNode head) {
ListNode fast=head;
ListNode low=head;
while(fast != null && fast.next != null){
low = low.next;
fast= fast.next.next;
}
return low;
}1.2.5 练习
//Linked List Cycle I（环形链表）
public boolean hasCycle(ListNode head) {
ListNode f=head;
ListNode s=head;
if(head == null || head.next == null) return false;
while(f!=null && f.next!=null){
f=f.next.next;
s=s.next;
if(f == s) return true;
}
return false;
}
//Merge k Sorted Lists（合并 k 个排序链表）
public ListNode mergeKLists(ListNode[] lists){
if(lists.length == 0)
return null;
if(lists.length == 1)
return lists[0];
if(lists.length == 2){
return mergeTwoLists(lists[0],lists[1]);
}
int mid = lists.length/2;
ListNode[] l1 = new ListNode[mid];
for(int i = 0; i < mid; i++){
l1[i] = lists[i];
}
ListNode[] l2 = new ListNode[lists.length-mid];
for(int i = mid,j=0; i < lists.length; i++,j++){
l2[j] = lists[i];
}
return mergeTwoLists(mergeKLists(l1),mergeKLists(l2));
}
public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
if (l1 == null) return l2;
if (l2 == null) return l1;
ListNode head = null;
if (l1.val <= l2.val){
head = l1;
head.next = mergeTwoLists(l1.next, l2);
} else {
head = l2;
head.next = mergeTwoLists(l1, l2.next);
}
return head;
}
