public class RedPacket {
/**
* 生成红包最小值 1分
*/
private static final int MIN_MONEY = 1;
/**
* 生成红包最大值 200人民币
*/
private static final int MAX_MONEY = 200 * 100;
/**
* 小于最小值
*/
private static final int LESS = -1;
/**
* 大于最大值
*/
private static final int MORE = -2;
/**
* 正常值
*/
private static final int OK = 1;
/**
* 最大的红包是平均值的 TIMES 倍，防止某一次分配红包较大
*/
private static final double TIMES = 2.1F;
private int recursiveCount = 0;
public List<Integer> splitRedPacket(int money, int count) {
List<Integer> moneys = new LinkedList<>();
//金额检查，如果最大红包 * 个数 < 总金额；则需要调大最小红包 MAX_MONEY
if (MAX_MONEY * count <= money) {
System.err.println("请调大最小红包金额 MAX_MONEY=[" + MAX_MONEY + "]");
return moneys ;
}
//计算出最大红包
int max = (int) ((money / count) * TIMES);
max = max > MAX_MONEY ? MAX_MONEY : max;
for (int i = 0; i < count; i++) {
//随机获取红包
int redPacket = randomRedPacket(money, MIN_MONEY, max, count - i);
moneys.add(redPacket);
//总金额每次减少
money -= redPacket;
}
return moneys;
}
private int randomRedPacket(int totalMoney, int minMoney, int maxMoney, int count) {
//只有一个红包直接返回
if (count == 1) {
return totalMoney;
}
if (minMoney == maxMoney) {
return minMoney;
}
//如果最大金额大于了剩余金额 则用剩余金额 因为这个 money 每分配一次都会减小
maxMoney = maxMoney > totalMoney ? totalMoney : maxMoney;
//在 minMoney到maxMoney 生成一个随机红包
int redPacket = (int) (Math.random() * (maxMoney - minMoney) + minMoney);
int lastMoney = totalMoney - redPacket;
int status = checkMoney(lastMoney, count - 1);
//正常金额
if (OK == status) {
return redPacket;
}
//如果生成的金额不合法 则递归重新生成
if (LESS == status) {
recursiveCount++;
System.out.println("recursiveCount==" + recursiveCount);
return randomRedPacket(totalMoney, minMoney, redPacket, count);
}
if (MORE == status) {
recursiveCount++;
System.out.println("recursiveCount===" + recursiveCount);
return randomRedPacket(totalMoney, redPacket, maxMoney, count);
}
return redPacket;
}
/**
* 校验剩余的金额的平均值是否在 最小值和最大值这个范围内
*
* @param lastMoney
* @param count
* @return
*/
private int checkMoney(int lastMoney, int count) {
double avg = lastMoney / count;
if (avg < MIN_MONEY) {
return LESS;
}
if (avg > MAX_MONEY) {
return MORE;
}
return OK;
}
public static void main(String[] args) {
RedPacket redPacket = new RedPacket();
List<Integer> redPackets = redPacket.splitRedPacket(20000, 100);
System.out.println(redPackets);
int sum = 0;
for (Integer red : redPackets) {
sum += red;
}
System.out.println(sum);
}
}public class BinaryNode {
private Object data ;
private BinaryNode left ;
private BinaryNode right ;
public BinaryNode() {
}
public BinaryNode(Object data, BinaryNode left, BinaryNode right) {
this.data = data;
this.left = left;
this.right = right;
}
public Object getData() {
return data;
}
public void setData(Object data) {
this.data = data;
}
public BinaryNode getLeft() {
return left;
}
public void setLeft(BinaryNode left) {
this.left = left;
}
public BinaryNode getRight() {
return right;
}
public void setRight(BinaryNode right) {
this.right = right;
}
public BinaryNode createNode(){
BinaryNode node = new BinaryNode("1",null,null) ;
BinaryNode left2 = new BinaryNode("2",null,null) ;
BinaryNode left3 = new BinaryNode("3",null,null) ;
BinaryNode left4 = new BinaryNode("4",null,null) ;
BinaryNode left5 = new BinaryNode("5",null,null) ;
BinaryNode left6 = new BinaryNode("6",null,null) ;
node.setLeft(left2) ;
left2.setLeft(left4);
left2.setRight(left6);
node.setRight(left3);
left3.setRight(left5) ;
return node ;
}
@Override
public String toString() {
return "BinaryNode{" +
"data=" + data +
", left=" + left +
", right=" + right +
'}';
}
/**
* 二叉树的层序遍历 借助于队列来实现 借助队列的先进先出的特性
*
* 首先将根节点入队列 然后遍历队列。
* 首先将根节点打印出来，接着判断左节点是否为空 不为空则加入队列
* @param node
*/
public void levelIterator(BinaryNode node){
LinkedList<BinaryNode> queue = new LinkedList<>() ;
//先将根节点入队
queue.offer(node) ;
BinaryNode current ;
while (!queue.isEmpty()){
current = queue.poll();
System.out.print(current.data+"--->");
if (current.getLeft() != null){
queue.offer(current.getLeft()) ;
}
if (current.getRight() != null){
queue.offer(current.getRight()) ;
}
}
}
}
public class BinaryNodeTest {
@Test
public void test1(){
BinaryNode node = new BinaryNode() ;
//创建二叉树
node = node.createNode() ;
System.out.println(node);
//层序遍历二叉树
node.levelIterator(node) ;
}
}/**
* Function: 判断一个数字是否为快乐数字 19 就是快乐数字 11就不是快乐数字
* 19
* 1*1+9*9=82
* 8*8+2*2=68
* 6*6+8*8=100
* 1*1+0*0+0*0=1
*
* 11
* 1*1+1*1=2
* 2*2=4
* 4*4=16
* 1*1+6*6=37
* 3*3+7*7=58
* 5*5+8*8=89
* 8*8+9*9=145
* 1*1+4*4+5*5=42
* 4*4+2*2=20
* 2*2+0*0=2
*
* 这里结果 1*1+1*1=2 和 2*2+0*0=2 重复，所以不是快乐数字
* @author crossoverJie
* Date: 04/01/2018 14:12
* @since JDK 1.8
*/
public class HappyNum {
/**
* 判断一个数字是否为快乐数字
* @param number
* @return
*/
public boolean isHappy(int number) {
Set<Integer> set = new HashSet<>(30);
while (number != 1) {
int sum = 0;
while (number > 0) {
//计算当前值的每位数的平方 相加的和 在放入set中，如果存在相同的就认为不是 happy数字
sum += (number % 10) * (number % 10);
number = number / 10;
}
if (set.contains(sum)) {
return false;
} else {
set.add(sum);
}
number = sum;
}
return true;
}
}
public class HappyNumTest {
@Test
public void isHappy() throws Exception {
HappyNum happyNum = new HappyNum() ;
boolean happy = happyNum.isHappy(19);
Assert.assertEquals(happy,true);
}
@Test
public void isHappy2() throws Exception {
HappyNum happyNum = new HappyNum() ;
boolean happy = happyNum.isHappy(11);
Assert.assertEquals(happy,false);
}
@Test
public void isHappy3() throws Exception {
HappyNum happyNum = new HappyNum() ;
boolean happy = happyNum.isHappy(100);
System.out.println(happy);
}
}/**
* Function:是否是环链表，采用快慢指针，一个走的快些一个走的慢些 如果最终相遇了就说明是环
* 就相当于在一个环形跑道里跑步，速度不一样的最终一定会相遇。
*
* @author crossoverJie
* Date: 04/01/2018 11:33
* @since JDK 1.8
*/
public class LinkLoop {
public static class Node{
private Object data ;
public Node next ;
public Node(Object data, Node next) {
this.data = data;
this.next = next;
}
public Node(Object data) {
this.data = data ;
}
}
/**
* 判断链表是否有环
* @param node
* @return
*/
public boolean isLoop(Node node){
Node slow = node ;
Node fast = node.next ;
while (slow.next != null){
Object dataSlow = slow.data;
Object dataFast = fast.data;
//说明有环
if (dataFast == dataSlow){
return true ;
}
//一共只有两个节点，但却不是环形链表的情况，判断NPE
if (fast.next == null){
return false ;
}
//slow走慢点 fast走快点
slow = slow.next ;
fast = fast.next.next ;
//如果走的快的发现为空 说明不存在环
if (fast == null){
return false ;
}
}
return false ;
}
}
public class LinkLoopTest {
/**
* 无环
* @throws Exception
*/
@Test
public void isLoop() throws Exception {
LinkLoop.Node node3 = new LinkLoop.Node("3");
LinkLoop.Node node2 = new LinkLoop.Node("2") ;
LinkLoop.Node node1 = new LinkLoop.Node("1") ;
node1.next = node2 ;
node2.next = node3 ;
LinkLoop linkLoop = new LinkLoop() ;
boolean loop = linkLoop.isLoop(node1);
Assert.assertEquals(loop,false);
}
/**
* 有环
* @throws Exception
*/
@Test
public void isLoop2() throws Exception {
LinkLoop.Node node3 = new LinkLoop.Node("3");
LinkLoop.Node node2 = new LinkLoop.Node("2") ;
LinkLoop.Node node1 = new LinkLoop.Node("1") ;
node1.next = node2 ;
node2.next = node3 ;
node3.next = node1 ;
LinkLoop linkLoop = new LinkLoop() ;
boolean loop = linkLoop.isLoop(node1);
Assert.assertEquals(loop,true);
}
/**
* 无环
* @throws Exception
*/
@Test
public void isLoop3() throws Exception {
LinkLoop.Node node2 = new LinkLoop.Node("2") ;
LinkLoop.Node node1 = new LinkLoop.Node("1") ;
node1.next = node2 ;
LinkLoop linkLoop = new LinkLoop() ;
boolean loop = linkLoop.isLoop(node1);
Assert.assertEquals(loop,false);
}
}/**
* Function:{1,3,5,7} target=8 返回{2,3}
*
* @author crossoverJie
* Date: 04/01/2018 09:53
* @since JDK 1.8
*/
public class TwoSum {
/**
* 时间复杂度为 O(N^2)
* @param nums
* @param target
* @return
*/
public int[] getTwo1(int[] nums,int target){
int[] result = new int[2] ;
for (int i= 0 ;i<nums.length ;i++){
int a = nums[i] ;
for (int j = nums.length -1 ;j >=0 ;j--){
int b = nums[j] ;
if ((a+b) == target){
result = new int[]{i,j} ;
}
}
}
return result ;
}
/**
* 时间复杂度 O(N)
* 利用Map Key存放目标值和当前值的差值，value 就是当前的下标
* 每次遍历是 查看当前遍历的值是否等于差值，如果是等于，说明两次相加就等于目标值。
* 然后取出 map 中 value ，和本次遍历的下标，就是两个下标值相加等于目标值了。
*
* @param nums
* @param target
* @return
*/
public int[] getTwo2(int[] nums,int target){
int[] result = new int[2] ;
Map<Integer,Integer> map = new HashMap<>(2) ;
for (int i=0 ;i<nums.length;i++){
if (map.containsKey(nums[i])){
result = new int[]{map.get(nums[i]),i} ;
}
map.put(target - nums[i],i) ;
}
return result ;
}
}
public class TwoSumTest {
@Test
public void getTwo1() throws Exception {
TwoSum twoSum = new TwoSum() ;
int[] nums ={1,3,5,7};
int[] two1 = twoSum.getTwo1(nums, 12);
System.out.println(JSON.toJSONString(two1));
}
@Test
public void getTwo2(){
TwoSum twoSum = new TwoSum() ;
int[] nums ={1,3,5,7};
int[] two = twoSum.getTwo2(nums, 10);
System.out.println(JSON.toJSONString(two));
}
}/**
* Function: 三种方式反向打印单向链表
*
* @author crossoverJie
* Date: 10/02/2018 16:14
* @since JDK 1.8
*/
public class ReverseNode {
/**
* 利用栈的先进后出特性
* @param node
*/
public void reverseNode1(Node node){
System.out.println("====翻转之前====");
Stack<Node> stack = new Stack<>() ;
while (node != null){
System.out.print(node.value + "===>");
stack.push(node) ;
node = node.next ;
}
System.out.println("");
System.out.println("====翻转之后====");
while (!stack.isEmpty()){
System.out.print(stack.pop().value + "===>");
}
}
/**
* 利用头插法插入链表
* @param head
*/
public void reverseNode(Node head) {
if (head == null) {
return ;
}
//最终翻转之后的 Node
Node node ;
Node pre = head;
Node cur = head.next;
Node next ;
while(cur != null){
next = cur.next;
//链表的头插法
cur.next = pre;
pre = cur;
cur = next;
}
head.next = null;
node = pre;
//遍历新链表
while (node != null){
System.out.println(node.value);
node = node.next ;
}
}
/**
* 递归
* @param node
*/
public void recNode(Node node){
if (node == null){
return ;
}
if (node.next != null){
recNode(node.next) ;
}
System.out.print(node.value+"===>");
}
public static class Node<T>{
public T value;
public Node<T> next ;
public Node(T value, Node<T> next ) {
this.next = next;
this.value = value;
}
}
}
//单测
public class ReverseNodeTest {
@Test
public void reverseNode1() throws Exception {
ReverseNode.Node<String> node4 = new Node<>("4",null) ;
Node<String> node3 = new Node<>("3",node4);
Node<String> node2 = new Node<>("2",node3);
Node<String> node1 = new Node("1",node2) ;
ReverseNode reverseNode = new ReverseNode() ;
reverseNode.reverseNode1(node1);
}
@Test
public void reverseNode12() throws Exception {
Node<String> node1 = new Node("1",null) ;
ReverseNode reverseNode = new ReverseNode() ;
reverseNode.reverseNode1(node1);
}
@Test
public void reverseNode13() throws Exception {
Node<String> node1 = null ;
ReverseNode reverseNode = new ReverseNode() ;
reverseNode.reverseNode1(node1);
}
/**
* 头插法
* @throws Exception
*/
@Test
public void reverseHead21() throws Exception {
Node<String> node4 = new Node<>("4",null) ;
Node<String> node3 = new Node<>("3",node4);
Node<String> node2 = new Node<>("2",node3);
Node<String> node1 = new Node("1",node2) ;
ReverseNode reverseNode = new ReverseNode() ;
reverseNode.reverseNode(node1);
}
@Test
public void recNodeTest31(){
Node<String> node4 = new Node<>("4",null) ;
Node<String> node3 = new Node<>("3",node4);
Node<String> node2 = new Node<>("2",node3);
Node<String> node1 = new Node("1",node2) ;
ReverseNode reverseNode = new ReverseNode() ;
reverseNode.recNode(node1);
}
}/**
* Function: 合并两个排好序的链表
*
* 每次比较两个链表的头结点，将较小结点放到新的链表，最后将新链表指向剩余的链表
*
* @author crossoverJie
* Date: 07/12/2017 13:58
* @since JDK 1.8
*/
public class MergeTwoSortedLists {
/**
* 1. 声明一个头结点
* 2. 将头结点的引用赋值给一个临时结点，也可以叫做下一结点。
* 3. 进行循环比较，每次都将指向值较小的那个结点(较小值的引用赋值给 lastNode )。
* 4. 再去掉较小值链表的头结点，指针后移。
* 5. lastNode 指针也向后移，由于 lastNode 是 head 的引用，这样可以保证最终 head 的值是往后更新的。
* 6. 当其中一个链表的指针移到最后时跳出循环。
* 7. 由于这两个链表已经是排好序的，所以剩下的链表必定是最大的值，只需要将指针指向它即可。
* 8. 由于 head 链表的第一个结点是初始化的0，所以只需要返回 0 的下一个结点即是合并了的链表。
* @param l1
* @param l2
* @return
*/
public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
ListNode head = new ListNode(0) ;
ListNode lastNode = head ;
while (l1 != null && l2 != null){
if (l1.currentVal < l2.currentVal){
lastNode.next = l1 ;
l1 = l1.next ;
} else {
lastNode.next = l2 ;
l2 = l2.next ;
}
lastNode =lastNode.next ;
}
if (l1 == null){
lastNode.next = l2 ;
}
if (l2 == null){
lastNode.next = l1 ;
}
return head.next ;
}
public static class ListNode {
/**
* 当前值
*/
int currentVal;
/**
* 下一个节点
*/
ListNode next;
ListNode(int val) {
currentVal = val;
}
@Override
public String toString() {
return "ListNode{" +
"currentVal=" + currentVal +
", next=" + next +
'}';
}
}
}
//单测
public class MergeTwoSortedListsTest {
MergeTwoSortedLists mergeTwoSortedLists ;
@Before
public void setUp() throws Exception {
mergeTwoSortedLists = new MergeTwoSortedLists();
}
@Test
public void mergeTwoLists() throws Exception {
ListNode l1 = new ListNode(1) ;
ListNode l1_2 = new ListNode(4);
l1.next = l1_2 ;
ListNode l1_3 = new ListNode(5) ;
l1_2.next = l1_3 ;
ListNode l2 = new ListNode(1) ;
ListNode l2_2 = new ListNode(3) ;
l2.next = l2_2 ;
ListNode l2_3 = new ListNode(6) ;
l2_2.next = l2_3 ;
ListNode l2_4 = new ListNode(9) ;
l2_3.next = l2_4 ;
ListNode listNode = mergeTwoSortedLists.mergeTwoLists(l1, l2);
ListNode node1 = new ListNode(1) ;
ListNode node2 = new ListNode(1);
node1.next = node2;
ListNode node3 = new ListNode(3) ;
node2.next= node3 ;
ListNode node4 = new ListNode(4) ;
node3.next = node4 ;
ListNode node5 = new ListNode(5) ;
node4.next = node5 ;
ListNode node6 = new ListNode(6) ;
node5.next = node6 ;
ListNode node7 = new ListNode(9) ;
node6.next = node7 ;
Assert.assertEquals(node1.toString(),listNode.toString());
}
@Test
public void mergeTwoLists2() throws Exception {
ListNode l2 = new ListNode(1) ;
ListNode l2_2 = new ListNode(3) ;
l2.next = l2_2 ;
ListNode l2_3 = new ListNode(6) ;
l2_2.next = l2_3 ;
ListNode l2_4 = new ListNode(9) ;
l2_3.next = l2_4 ;
ListNode listNode = mergeTwoSortedLists.mergeTwoLists(null, l2);
System.out.println(listNode.toString());
}
@Test
public void mergeTwoLists3() throws Exception {
ListNode l2 = new ListNode(1) ;
ListNode l2_2 = new ListNode(3) ;
l2.next = l2_2 ;
ListNode l2_3 = new ListNode(6) ;
l2_2.next = l2_3 ;
ListNode l2_4 = new ListNode(9) ;
l2_3.next = l2_4 ;
ListNode listNode = mergeTwoSortedLists.mergeTwoLists(l2, null);
ListNode node1 = new ListNode(1) ;
ListNode node2 = new ListNode(3);
node1.next = node2;
ListNode node3 = new ListNode(6) ;
node2.next= node3 ;
ListNode node4 = new ListNode(9) ;
node3.next = node4 ;
Assert.assertEquals(node1.toString(),listNode.toString());
}
}/**
* Function: 两个栈实现队列
*
* 利用两个栈来实现，第一个栈存放写队列的数据。
* 第二个栈存放移除队列的数据，移除之前先判断第二个栈里是否有数据。
* 如果没有就要将第一个栈里的数据依次弹出压入第二个栈，这样写入之后的顺序再弹出其实就是一个先进先出的结构了。
*
* 这样出队列只需要移除第二个栈的头元素即可。
*
* @author crossoverJie
* Date: 09/02/2018 23:51
* @since JDK 1.8
*/
public class TwoStackQueue<T> {
/**
* 写入的栈
*/
private Stack<T> input = new Stack() ;
/**
* 移除队列所出的栈
*/
private Stack<T> out = new Stack() ;
/**
* 写入队列
* @param t
*/
public void appendTail(T t){
input.push(t) ;
}
/**
* 删除队列头结点 并返回删除数据
* @return
*/
public T deleteHead(){
//是空的 需要将 input 出栈写入 out
if (out.isEmpty()){
while (!input.isEmpty()){
out.push(input.pop()) ;
}
}
//不为空时直接移除出栈就表示移除了头结点
return out.pop() ;
}
public int getSize(){
return input.size() + out.size() ;
}
}
//单测
public class TwoStackQueueTest {
private final static Logger LOGGER = LoggerFactory.getLogger(TwoStackQueueTest.class);
@Test
public void queue(){
TwoStackQueue<String> twoStackQueue = new TwoStackQueue<String>() ;
twoStackQueue.appendTail("1") ;
twoStackQueue.appendTail("2") ;
twoStackQueue.appendTail("3") ;
twoStackQueue.appendTail("4") ;
twoStackQueue.appendTail("5") ;
int size = twoStackQueue.getSize();
for (int i = 0; i< size ; i++){
LOGGER.info(twoStackQueue.deleteHead());
}
LOGGER.info("========第二次添加=========");
twoStackQueue.appendTail("6") ;
size = twoStackQueue.getSize();
for (int i = 0; i< size ; i++){
LOGGER.info(twoStackQueue.deleteHead());
}
}
}/**
* 链表排序, 建议使用归并排序，
* 问题描述，给定一个Int的链表，要求在时间最优的情况下完成链表元素由大到小的排序，
* e.g: 1->5->4->3->2
* 排序后结果 5->4->3->2->1
*
* @author 6563699600@qq.com
* @date 6/7/2018 11:42 PM
* @since 1.0
*/
public class LinkedListMergeSort {
/**
* 定义链表数据结构，包含当前元素，以及当前元素的后续元素指针
*/
final static class Node {
int e;
Node next;
public Node() {
}
public Node(int e, Node next) {
this.e = e;
this.next = next;
}
}
public Node mergeSort(Node first, int length) {
if (length == 1) {
return first;
} else {
Node middle = new Node();
Node tmp = first;
/**
* 后期会对这里进行优化，通过一次遍历算出长度和中间元素
*/
for (int i = 0; i < length; i++) {
if (i == length / 2) {
break;
}
middle = tmp;
tmp = tmp.next;
}
/**
* 这里是链表归并时要注意的细节
* 在链表进行归并排序过程中，会涉及到将一个链表打散为两个独立的链表，所以需要在中间元素的位置将其后续指针指为null；
*/
Node right = middle.next;
middle.next = null;
Node leftStart = mergeSort(first, length / 2);
Node rightStart;
if (length % 2 == 0) {
rightStart = mergeSort(right, length / 2);
} else {
rightStart = mergeSort(right, length / 2 + 1);
}
return mergeList(leftStart, rightStart);
}
}
/**
* 合并链表，具体的实现细节可参考<code>MergeTwoSortedLists</code>
*
* @param left
* @param right
* @return
*/
public Node mergeList(Node left, Node right) {
Node head = new Node();
Node result = head;
/**
* 思想就是两个链表同时遍历，将更的元素插入结果中，同时更更大的元素所属的链表的指针向下移动
*/
while (!(null == left && null == right)) {
Node tmp;
if (left == null) {
result.next = right;
break;
} else if (right == null) {
result.next = left;
break;
} else if (left.e >= right.e) {
tmp = left;
result.next = left;
result = tmp;
left = left.next;
} else {
tmp = right;
result.next = right;
result = tmp;
right = right.next;
}
}
return head.next;
}
public static void main(String[] args) {
Node head = new Node();
head.next = new Node(7,
new Node(2,
new Node(5,
new Node(4,
new Node(3,
new Node(6,
new Node(11, null)
)
)
)
)
)
);
int length = 0;
for (Node e = head.next; null != e; e = e.next) {
length++;
}
LinkedListMergeSort sort = new LinkedListMergeSort();
head.next = sort.mergeSort(head.next, length);
for (Node n = head.next; n != null; n = n.next) {
System.out.println(n.e);
}
}
}/**
* 数组右移K次, 原数组<code> [1, 2, 3, 4, 5, 6, 7]</code> 右移3次后结果为 <code>[5,6,7,1,2,3,4]</code>
*
* 基本思路：不开辟新的数组空间的情况下考虑在原属组上进行操作
* 1 将数组倒置，这样后k个元素就跑到了数组的前面，然后反转一下即可
* 2 同理后 len-k个元素只需要翻转就完成数组的k次移动
*
* @author 656369960@qq.com
* @date 12/7/2018 1:38 PM
* @since 1.0
*/
public class ArrayKShift {
public void arrayKShift(int[] array, int k) {
/**
* constrictions
*/
if (array == null || 0 == array.length) {
return ;
}
k = k % array.length;
if (0 > k) {
return;
}
/**
* reverse array , e.g: [1, 2, 3 ,4] to [4,3,2,1]
*/
for (int i = 0; i < array.length / 2; i++) {
int tmp = array[i];
array[i] = array[array.length - 1 - i];
array[array.length - 1 - i] = tmp;
}
/**
* first k element reverse
*/
for (int i = 0; i < k / 2; i++) {
int tmp = array[i];
array[i] = array[k - 1 - i];
array[k - 1 - i] = tmp;
}
/**
* last length - k element reverse
*/
for (int i = k; i < k + (array.length - k ) / 2; i ++) {
int tmp = array[i];
array[i] = array[array.length - 1 - i + k];
array[array.length - 1 - i + k] = tmp;
}
}
public static void main(String[] args) {
int[] array = {1, 2, 3 ,4, 5, 6, 7};
ArrayKShift shift = new ArrayKShift();
shift.arrayKShift(array, 6);
Arrays.stream(array).forEach(o -> {
System.out.println(o);
});
}
}/**
* Function: 两个线程交替执行打印 1~100
*
* lock 版
*
* @author crossoverJie
* Date: 11/02/2018 10:04
* @since JDK 1.8
*/
public class TwoThread {
private int start = 1;
/**
* 对 flag 的写入虽然加锁保证了线程安全，但读取的时候由于 不是 volatile 所以可能会读取到旧值
*
*/
private volatile boolean flag = false;
/**
* 重入锁
*/
private final static Lock LOCK = new ReentrantLock();
public static void main(String[] args) {
TwoThread twoThread = new TwoThread();
Thread t1 = new Thread(new OuNum(twoThread));
t1.setName("t1");
Thread t2 = new Thread(new JiNum(twoThread));
t2.setName("t2");
t1.start();
t2.start();
}
/**
* 偶数线程
*/
public static class OuNum implements Runnable {
private TwoThread number;
public OuNum(TwoThread number) {
this.number = number;
}
@Override
public void run() {
while (number.start <= 1000) {
if (number.flag) {
try {
LOCK.lock();
System.out.println(Thread.currentThread().getName() + "+-+" + number.start);
number.start++;
number.flag = false;
} finally {
LOCK.unlock();
}
}
}
}
}
/**
* 奇数线程
*/
public static class JiNum implements Runnable {
private TwoThread number;
public JiNum(TwoThread number) {
this.number = number;
}
@Override
public void run() {
while (number.start <= 1000) {
if (!number.flag) {
try {
LOCK.lock();
System.out.println(Thread.currentThread().getName() + "+-+" + number.start);
number.start++;
number.flag = true;
} finally {
LOCK.unlock();
}
}
}
}
}
}/**
* Function:两个线程交替执行打印 1~100
* 等待通知机制版
*
* @author crossoverJie
* Date: 07/03/2018 13:19
* @since JDK 1.8
*/
public class TwoThreadWaitNotify {
private int start = 1;
private boolean flag = false;
public static void main(String[] args) {
TwoThreadWaitNotify twoThread = new TwoThreadWaitNotify();
Thread t1 = new Thread(new OuNum(twoThread));
t1.setName("t1");
Thread t2 = new Thread(new JiNum(twoThread));
t2.setName("t2");
t1.start();
t2.start();
}
/**
* 偶数线程
*/
public static class OuNum implements Runnable {
private TwoThreadWaitNotify number;
public OuNum(TwoThreadWaitNotify number) {
this.number = number;
}
@Override
public void run() {
while (number.start <= 100) {
synchronized (TwoThreadWaitNotify.class) {
System.out.println("偶数线程抢到锁了");
if (number.flag) {
System.out.println(Thread.currentThread().getName() + "+-+偶数" + number.start);
number.start++;
number.flag = false;
TwoThreadWaitNotify.class.notify();
}else {
try {
TwoThreadWaitNotify.class.wait();
} catch (InterruptedException e) {
e.printStackTrace();
}
}
}
}
}
}
/**
* 奇数线程
*/
public static class JiNum implements Runnable {
private TwoThreadWaitNotify number;
public JiNum(TwoThreadWaitNotify number) {
this.number = number;
}
@Override
public void run() {
while (number.start <= 100) {
synchronized (TwoThreadWaitNotify.class) {
System.out.println("奇数线程抢到锁了");
if (!number.flag) {
System.out.println(Thread.currentThread().getName() + "+-+奇数" + number.start);
number.start++;
number.flag = true;
TwoThreadWaitNotify.class.notify();
}else {
try {
TwoThreadWaitNotify.class.wait();
} catch (InterruptedException e) {
e.printStackTrace();
}
}
}
}
}
}
}/**
* Function: 两个线程交替执行打印 1~100
* <p>
* non blocking 版：
* 两个线程轮询volatile变量(flag)
* 线程一"看到"flag值为1时执行代码并将flag设置为0,
* 线程二"看到"flag值为0时执行代码并将flag设置未1,
* 2个线程不断轮询直到满足条件退出
*
* @author twoyao
* Date: 05/07/2018
* @since JDK 1.8
*/
public class TwoThreadNonBlocking implements Runnable {
/**
* 当flag为1时只有奇数线程可以执行，并将其置为0
* 当flag为0时只有偶数线程可以执行，并将其置为1
*/
private volatile static int flag = 1;
private int start;
private int end;
private String name;
private TwoThreadNonBlocking(int start, int end, String name) {
this.name = name;
this.start = start;
this.end = end;
}
@Override
public void run() {
while (start <= end) {
int f = flag;
if ((start & 0x01) == f) {
System.out.println(name + "+-+" + start);
start += 2;
// 因为只可能同时存在一个线程修改该值，所以不会存在竞争
flag ^= 0x1;
}
}
}
public static void main(String[] args) {
new Thread(new TwoThreadNonBlocking(1, 100, "t1")).start();
new Thread(new TwoThreadNonBlocking(2, 100, "t2")).start();
}
}