《算法笔记》2. 链表、栈、队列、递归、哈希表、顺序表

1 链表、栈、队列、递归、哈希

转载注明出处,源码地址: https://github.com/Dairongpeng/algorithm-note ,欢迎star

1.1 链表

1.1.1 单向链表

单向链表的节点结构(可以实现成泛型) :

    public class Node {
        public int value;
        public Node next;
        
        public Node(int data) {
            value = data;
        }
        
    }

1.1.2 双向链表

双向链表的节点结构(可以实现成功泛型):

	public static class DoubleNode {
		public int value;
		public DoubleNode last;
		public DoubleNode next;

		public DoubleNode(int data) {
			value = data;
		}
	}

1.1.3 单双链表简单练习

  1. 单链表和双链表如何反转

1 -> 2 -> 3 转换为 3 -> 2 -> 1


package class02;

import java.util.ArrayList;

public class Code01_ReverseList {

	public static class Node {
		public int value;
		public Node next;

		public Node(int data) {
			value = data;
		}
	}

	public static class DoubleNode {
		public int value;
		public DoubleNode last;
		public DoubleNode next;

		public DoubleNode(int data) {
			value = data;
		}
	}

    // 翻转单向链表,传入头结点
	public static Node reverseLinkedList(Node head) {
		Node pre = null;
		Node next = null;
		while (head != null) {
			next = head.next;
			head.next = pre;
			pre = head;
			head = next;
		}
		return pre;
	}

    // 翻转双向链表,传入头结点
	public static DoubleNode reverseDoubleList(DoubleNode head) {
		DoubleNode pre = null;
		DoubleNode next = null;
		while (head != null) {
			next = head.next;
			head.next = pre;
			head.last = next;
			pre = head;
			head = next;
		}
		return pre;
	}

	public static Node testReverseLinkedList(Node head) {
		if (head == null) {
			return null;
		}
		ArrayList<Node> list = new ArrayList<>();
		while (head != null) {
			list.add(head);
			head = head.next;
		}
		list.get(0).next = null;
		int N = list.size();
		for (int i = 1; i < N; i++) {
			list.get(i).next = list.get(i - 1);
		}
		return list.get(N - 1);
	}

	public static DoubleNode testReverseDoubleList(DoubleNode head) {
		if (head == null) {
			return null;
		}
		ArrayList<DoubleNode> list = new ArrayList<>();
		while (head != null) {
			list.add(head);
			head = head.next;
		}
		list.get(0).next = null;
		DoubleNode pre = list.get(0);
		int N = list.size();
		for (int i = 1; i < N; i++) {
			DoubleNode cur = list.get(i);
			cur.last = null;
			cur.next = pre;
			pre.last = cur;
			pre = cur;
		}
		return list.get(N - 1);
	}

	public static Node generateRandomLinkedList(int len, int value) {
		int size = (int) (Math.random() * (len + 1));
		if (size == 0) {
			return null;
		}
		size--;
		Node head = new Node((int) (Math.random() * (value + 1)));
		Node pre = head;
		while (size != 0) {
			Node cur = new Node((int) (Math.random() * (value + 1)));
			pre.next = cur;
			pre = cur;
			size--;
		}
		return head;
	}

	public static DoubleNode generateRandomDoubleList(int len, int value) {
		int size = (int) (Math.random() * (len + 1));
		if (size == 0) {
			return null;
		}
		size--;
		DoubleNode head = new DoubleNode((int) (Math.random() * (value + 1)));
		DoubleNode pre = head;
		while (size != 0) {
			DoubleNode cur = new DoubleNode((int) (Math.random() * (value + 1)));
			pre.next = cur;
			cur.last = pre;
			pre = cur;
			size--;
		}
		return head;
	}

	// 要求无环,有环别用这个函数
	public static boolean checkLinkedListEqual(Node head1, Node head2) {
		while (head1 != null && head2 != null) {
			if (head1.value != head2.value) {
				return false;
			}
			head1 = head1.next;
			head2 = head2.next;
		}
		return head1 == null && head2 == null;
	}

	// 要求无环,有环别用这个函数
	public static boolean checkDoubleListEqual(DoubleNode head1, DoubleNode head2) {
		boolean null1 = head1 == null;
		boolean null2 = head2 == null;
		if (null1 && null2) {
			return true;
		}
		if (null1 ^ null2) {
			return false;
		}
		if (head1.last != null || head2.last != null) {
			return false;
		}
		DoubleNode end1 = null;
		DoubleNode end2 = null;
		while (head1 != null && head2 != null) {
			if (head1.value != head2.value) {
				return false;
			}
			end1 = head1;
			end2 = head2;
			head1 = head1.next;
			head2 = head2.next;
		}
		if (head1 != null || head2 != null) {
			return false;
		}
		while (end1 != null && end2 != null) {
			if (end1.value != end2.value) {
				return false;
			}
			end1 = end1.last;
			end2 = end2.last;
		}
		return end1 == null && end2 == null;
	}

	public static void main(String[] args) {
		int len = 50;
		int value = 100;
		int testTime = 100000;
		for (int i = 0; i < testTime; i++) {
			Node node1 = generateRandomLinkedList(len, value);
			Node reverse1 = reverseLinkedList(node1);
			Node back1 = testReverseLinkedList(reverse1);
			if (!checkLinkedListEqual(node1, back1)) {
				System.out.println("oops!");
				break;
			}
			DoubleNode node2 = generateRandomDoubleList(len, value);
			DoubleNode reverse2 = reverseDoubleList(node2);
			DoubleNode back2 = testReverseDoubleList(reverse2);
			if (!checkDoubleListEqual(node2, back2)) {
				System.out.println("oops!");
				break;
			}
		}
		System.out.println("finish!");

	}

}

  1. 把给定的值都删除

比如给定一个链表头结点,删除该节点上值为3的节点,那么可能头结点就是3,存在删头部的情况,这里最终返回应该是删除所有值为3的节点之后的新的头部


package class02;

public class Code02_DeleteGivenValue {

	public static class Node {
		public int value;
		public Node next;

		public Node(int data) {
			this.value = data;
		}
	}

    // 先检查头部,寻找第一个不等于需要删除的值的节点,就是新的头部
	public static Node removeValue(Node head, int num) {
		while (head != null) {
			if (head.value != num) {
				break;
			}
			head = head.next;
		}
		// head来到 第一个不需要删的位置
		Node pre = head;
		Node cur = head;
		// 
		while (cur != null) {
			if (cur.value == num) {
				pre.next = cur.next;
			} else {
				pre = cur;
			}
			cur = cur.next;
		}
		return head;
	}

}

Tips: Java中也有可能产生内存泄漏,与CPP不同,CPP的内存泄漏有可能是我们开辟了内存空间忘记释放。而Java的内存泄漏大可能是程序中的变量的生存周期引起的,如果该程序是一个类似定时任务的7*24小时不间断运行,那么申请的变量(数据结构)就有可能不会被及时释放。如果不注意往里面添加一些不必要的变量,这些变量就是内存泄漏

1.2 栈、队列

  1. 逻辑概念

栈:数据先进后出,犹如弹夹

队列: 数据先进先出,排队

  1. 底层实现方式

双向链表实现

package class02;

import java.util.LinkedList;
import java.util.Queue;
import java.util.Stack;

public class Code03_DoubleEndsQueueToStackAndQueue {

	public static class Node<T> {
		public T value;
		public Node<T> last;
		public Node<T> next;

		public Node(T data) {
			value = data;
		}
	}

	public static class DoubleEndsQueue<T> {
		public Node<T> head;
		public Node<T> tail;

        // 从头部加节点
		public void addFromHead(T value) {
			Node<T> cur = new Node<T>(value);
			if (head == null) {
				head = cur;
				tail = cur;
			} else {
				cur.next = head;
				head.last = cur;
				head = cur;
			}
		}

        // 从尾部加节点
		public void addFromBottom(T value) {
			Node<T> cur = new Node<T>(value);
			if (head == null) {
				head = cur;
				tail = cur;
			} else {
				cur.last = tail;
				tail.next = cur;
				tail = cur;
			}
		}

        // 从头部弹出节点
		public T popFromHead() {
			if (head == null) {
				return null;
			}
			Node<T> cur = head;
			if (head == tail) {
				head = null;
				tail = null;
			} else {
				head = head.next;
				cur.next = null;
				head.last = null;
			}
			return cur.value;
		}

        // 从尾部弹出节点
		public T popFromBottom() {
			if (head == null) {
				return null;
			}
			Node<T> cur = tail;
			if (head == tail) {
				head = null;
				tail = null;
			} else {
				tail = tail.last;
				tail.next = null;
				cur.last = null;
			}
			return cur.value;
		}

        // 该双向链表结构是否为空
		public boolean isEmpty() {
			return head == null;
		}

	}

    // 用上述双向链表结构实现栈
	public static class MyStack<T> {
		private DoubleEndsQueue<T> queue;

		public MyStack() {
			queue = new DoubleEndsQueue<T>();
		}

		public void push(T value) {
			queue.addFromHead(value);
		}

		public T pop() {
			return queue.popFromHead();
		}

		public boolean isEmpty() {
			return queue.isEmpty();
		}

	}

    // 用上述双向链表结构实现队列
	public static class MyQueue<T> {
		private DoubleEndsQueue<T> queue;

		public MyQueue() {
			queue = new DoubleEndsQueue<T>();
		}

		public void push(T value) {
			queue.addFromHead(value);
		}

		public T poll() {
			return queue.popFromBottom();
		}

		public boolean isEmpty() {
			return queue.isEmpty();
		}

	}

	public static boolean isEqual(Integer o1, Integer o2) {
		if (o1 == null && o2 != null) {
			return false;
		}
		if (o1 != null && o2 == null) {
			return false;
		}
		if (o1 == null && o2 == null) {
			return true;
		}
		return o1.equals(o2);
	}

	public static void main(String[] args) {
		int oneTestDataNum = 100;
		int value = 10000;
		int testTimes = 100000;
		for (int i = 0; i < testTimes; i++) {
			MyStack<Integer> myStack = new MyStack<>();
			MyQueue<Integer> myQueue = new MyQueue<>();
			Stack<Integer> stack = new Stack<>();
			Queue<Integer> queue = new LinkedList<>();
			for (int j = 0; j < oneTestDataNum; j++) {
				int nums = (int) (Math.random() * value);
				if (stack.isEmpty()) {
					myStack.push(nums);
					stack.push(nums);
				} else {
					if (Math.random() < 0.5) {
						myStack.push(nums);
						stack.push(nums);
					} else {
						if (!isEqual(myStack.pop(), stack.pop())) {
							System.out.println("oops!");
						}
					}
				}
				int numq = (int) (Math.random() * value);
				if (stack.isEmpty()) {
					myQueue.push(numq);
					queue.offer(numq);
				} else {
					if (Math.random() < 0.5) {
						myQueue.push(numq);
						queue.offer(numq);
					} else {
						if (!isEqual(myQueue.poll(), queue.poll())) {
							System.out.println("oops!");
						}
					}
				}
			}
		}
		System.out.println("finish!");
	}

}

数组实现,对于栈特别简单,对于队列,如下

package class02;

public class Code04_RingArray {

	public static class MyQueue {
	    // 数组结构
		private int[] arr;
		// 往当前队列添加数的下标位置
		private int pushi;
		// 当前队列需要出队列的位置
		private int polli;
		// 当前队列使用的空间大小
		private int size;
		// 数组最大大小,用户传入
		private final int limit;

		public MyQueue(int limit) {
			arr = new int[limit];
			pushi = 0;
			polli = 0;
			size = 0;
			this.limit = limit;
		}

		public void push(int value) {
			if (size == limit) {
				throw new RuntimeException("栈满了,不能再加了");
			}
			size++;
			arr[pushi] = value;
			pushi = nextIndex(pushi);
		}

		public int pop() {
			if (size == 0) {
				throw new RuntimeException("栈空了,不能再拿了");
			}
			size--;
			int ans = arr[polli];
			polli = nextIndex(polli);
			return ans;
		}

		public boolean isEmpty() {
			return size == 0;
		}

		// 如果现在的下标是i,返回下一个位置,该实现可以实现环形的ringbuffer
		private int nextIndex(int i) {
			return i < limit - 1 ? i + 1 : 0;
		}

	}

}

1.3 栈、队列常见面试题

一、实现一个特殊的栈,在基本功能的基础上,再实现返回栈中最小元素的功能更

1、pop、push、getMin操作的时间复杂度都是O(1)

2、设计的栈类型可以使用现成的栈结构

思路:准备两个栈,一个data栈,一个min栈。数据压data栈,min栈对比min栈顶元素,谁小加谁。这样的话data栈和min栈是同步上升的,元素个数一样多,且min栈的栈顶,是data栈所有元素中最小的那个。数据弹出data栈,我们同步弹出min栈,保证个数相等,切min栈弹出的就是最小值

package class02;

import java.util.Stack;

public class Code05_GetMinStack {

	public static class MyStack1 {
		private Stack<Integer> stackData;
		private Stack<Integer> stackMin;

		public MyStack1() {
			this.stackData = new Stack<Integer>();
			this.stackMin = new Stack<Integer>();
		}

		public void push(int newNum) {
		    // 当前最小栈为空,直接压入
			if (this.stackMin.isEmpty()) {
				this.stackMin.push(newNum);
			// 当前元素小于最小栈的栈顶,压入当前值
			} else if (newNum <= this.getmin()) {
				this.stackMin.push(newNum);
			}
			// 往数据栈中压入当前元素
			this.stackData.push(newNum);
		}

		public int pop() {
			if (this.stackData.isEmpty()) {
				throw new RuntimeException("Your stack is empty.");
			}
			int value = this.stackData.pop();
			if (value == this.getmin()) {
				this.stackMin.pop();
			}
			return value;
		}

		public int getmin() {
			if (this.stackMin.isEmpty()) {
				throw new RuntimeException("Your stack is empty.");
			}
			return this.stackMin.peek();
		}
	}

	public static class MyStack2 {
		private Stack<Integer> stackData;
		private Stack<Integer> stackMin;

		public MyStack2() {
			this.stackData = new Stack<Integer>();
			this.stackMin = new Stack<Integer>();
		}

		public void push(int newNum) {
			if (this.stackMin.isEmpty()) {
				this.stackMin.push(newNum);
			} else if (newNum < this.getmin()) {
				this.stackMin.push(newNum);
			} else {
				int newMin = this.stackMin.peek();
				this.stackMin.push(newMin);
			}
			this.stackData.push(newNum);
		}

		public int pop() {
			if (this.stackData.isEmpty()) {
				throw new RuntimeException("Your stack is empty.");
			}
			// 弹出操作,同步弹出,保证大小一致,只返回给用户data栈中的内容即可
			this.stackMin.pop();
			return this.stackData.pop();
		}

		public int getmin() {
			if (this.stackMin.isEmpty()) {
				throw new RuntimeException("Your stack is empty.");
			}
			return this.stackMin.peek();
		}
	}

	public static void main(String[] args) {
		MyStack1 stack1 = new MyStack1();
		stack1.push(3);
		System.out.println(stack1.getmin());
		stack1.push(4);
		System.out.println(stack1.getmin());
		stack1.push(1);
		System.out.println(stack1.getmin());
		System.out.println(stack1.pop());
		System.out.println(stack1.getmin());

		System.out.println("=============");

		MyStack1 stack2 = new MyStack1();
		stack2.push(3);
		System.out.println(stack2.getmin());
		stack2.push(4);
		System.out.println(stack2.getmin());
		stack2.push(1);
		System.out.println(stack2.getmin());
		System.out.println(stack2.pop());
		System.out.println(stack2.getmin());
	}

}

二、如何用栈结构实现队列结构,如何用队列结构实现栈结构

这两种结构的应用实在太多,刷题时会大量见到

/**
* 两个栈实现队列
**/
package class02;

import java.util.Stack;

public class Code06_TwoStacksImplementQueue {

	public static class TwoStacksQueue {
		public Stack<Integer> stackPush;
		public Stack<Integer> stackPop;

		public TwoStacksQueue() {
			stackPush = new Stack<Integer>();
			stackPop = new Stack<Integer>();
		}

		// push栈向pop栈倒入数据
		private void pushToPop() {
			if (stackPop.empty()) {
				while (!stackPush.empty()) {
					stackPop.push(stackPush.pop());
				}
			}
		}

		public void add(int pushInt) {
			stackPush.push(pushInt);
			pushToPop();
		}

		public int poll() {
			if (stackPop.empty() && stackPush.empty()) {
				throw new RuntimeException("Queue is empty!");
			}
			pushToPop();
			return stackPop.pop();
		}

		public int peek() {
			if (stackPop.empty() && stackPush.empty()) {
				throw new RuntimeException("Queue is empty!");
			}
			pushToPop();
			return stackPop.peek();
		}
	}

	public static void main(String[] args) {
		TwoStacksQueue test = new TwoStacksQueue();
		test.add(1);
		test.add(2);
		test.add(3);
		System.out.println(test.peek());
		System.out.println(test.poll());
		System.out.println(test.peek());
		System.out.println(test.poll());
		System.out.println(test.peek());
		System.out.println(test.poll());
	}

}

/**
* 两个队列实现栈 
**/
package class02;

import java.util.LinkedList;
import java.util.Queue;
import java.util.Stack;

public class Code07_TwoQueueImplementStack {

	public static class TwoQueueStack<T> {
		public Queue<T> queue;
		public Queue<T> help;

		public TwoQueueStack() {
			queue = new LinkedList<>();
			help = new LinkedList<>();
		}

		public void push(T value) {
			queue.offer(value);
		}

		public T poll() {
			while (queue.size() > 1) {
				help.offer(queue.poll());
			}
			T ans = queue.poll();
			Queue<T> tmp = queue;
			queue = help;
			help = tmp;
			return ans;
		}

		public T peek() {
			while (queue.size() > 1) {
				help.offer(queue.poll());
			}
			T ans = queue.poll();
			help.offer(ans);
			Queue<T> tmp = queue;
			queue = help;
			help = tmp;
			return ans;
		}

		public boolean isEmpty() {
			return queue.isEmpty();
		}

	}

	public static void main(String[] args) {
		System.out.println("test begin");
		TwoQueueStack<Integer> myStack = new TwoQueueStack<>();
		Stack<Integer> test = new Stack<>();
		int testTime = 1000000;
		int max = 1000000;
		for (int i = 0; i < testTime; i++) {
			if (myStack.isEmpty()) {
				if (!test.isEmpty()) {
					System.out.println("Oops");
				}
				int num = (int) (Math.random() * max);
				myStack.push(num);
				test.push(num);
			} else {
				if (Math.random() < 0.25) {
					int num = (int) (Math.random() * max);
					myStack.push(num);
					test.push(num);
				} else if (Math.random() < 0.5) {
					if (!myStack.peek().equals(test.peek())) {
						System.out.println("Oops");
					}
				} else if (Math.random() < 0.75) {
					if (!myStack.poll().equals(test.pop())) {
						System.out.println("Oops");
					}
				} else {
					if (myStack.isEmpty() != test.isEmpty()) {
						System.out.println("Oops");
					}
				}
			}
		}

		System.out.println("test finish!");

	}

}

1.4 递归

1、从思想上理解递归

2、从实现角度出发理解递归

例子:

求数组arr[L...R]中的最大值,怎么用递归方法实现

1、 将[L...R]范围分成左右两半。左[L...Mid],右[Mid+1...R]
2、 左部分求最大值,右部分求最大值
3、[L...R]范围上的最大值,就是max{左部分最大值,右部分最大值}

2步骤是个递归过程,当范围上只有一个数,就可以不用再递归了

package class02;

public class Code08_GetMax {

	// 求arr中的最大值
	public static int getMax(int[] arr) {
		return process(arr, 0, arr.length - 1);
	}

	// arr[L..R]范围上求最大值  L ... R   N
	public static int process(int[] arr, int L, int R) {
		if (L == R) { // arr[L..R]范围上只有一个数,直接返回,base case
			return arr[L];
		}
		int mid = L + ((R - L) >> 1); // 中点
		// 左部分最大值
		int leftMax = process(arr, L, mid);
		// 右部分最大值
		int rightMax = process(arr, mid + 1, R);
		return Math.max(leftMax, rightMax);
	}

}

递归在系统中是怎么实现的?递归实际上利用的是系统栈来实现的。保存当前调用现场,去执行子问题,子问题的返回作为现场的需要的参数填充,最终构建还原栈顶的现场,返回。所以递归行为不是玄学,任何递归都可以改为非递归实现,我们自己压栈用迭代等实现就行

1.4.1 递归行为的时间复杂度

对于满足

T(N) = aT(N/b) + O(N^d)

其中: a,b,d为常数

公式表示,子问题的规模是一致的,该子问题调用了a次,N/b代表子问题的规模,O(N^d)为除去递归调用剩余的时间复杂度。

比如上述问题的递归,[L...R]上有N个数,第一个子问题的规模是N/2,第二个子问题的规模也是N/2。子问题调用了2次。额为复杂度为O(1),那么公式为:

T(N) = 2T(N/2) + O(N^0)

结论:如果我们的递归满足这种公式,那么该递归的时间复杂度(Master公式)为

logb^a > d   =>  O(N ^ (logb^a))

logb^a < d   =>  O(N^d)

logb^a == d   =>  O(N^d * logN)

那么上述问题的a=2, b=2,d=0,满足第一条,递归时间复杂度为:O(N)

1.5 哈希表HashMap、HashSet

Hash表的增删改查,在使用的时候,一律认为时间复杂度是O(1)的

在Java中,int double float基础类型,按值传递; Integer, Double, Float按引用传递的,比较包装类型的值是否相等,使用equals方法。

注意:在Java底层,包装类如果范围比较小,底层仍然采用值传递,比如Integer如果范围在-128~127之间,是按值传递的

但是在Hash表中,即使是包装类型的key,我们也一律按值传递,例如Hash<Integer,String>如果我们put相同的key的值,那么不会产生两个值相等的key而是覆盖操作。但是Hash表并不是一直是按值传递的,只是针对包装类型,如果是我们自定义的引用类型,那么仍然按引用传递

1.6 顺序表 TreeMap、TreeSet

顺序表比哈希表功能多,但是顺序表的很多操作时间复杂度是O(logN)

有序表的底层可以有很多结构实现,比如AVL树,SB树,红黑树,跳表。其中AVL,SB,红黑都是具备各自平衡性的搜索二叉树

由于平衡二叉树每时每刻都会维持自身的平衡,所以操作为O(logN)。暂时理解,后面会单独整理

由于满足去重排序功能来维持底层树的平衡,所以如果是基础类型和包装类型的key直接按值来做比较,但是如果我们的key是自己定义的类型,那么我们要自己制定比较规则(比较器),用来让底层的树保持比较后的平衡

package class02;

import java.util.HashMap;
import java.util.HashSet;
import java.util.TreeMap;

public class HashMapAndSortedMap {
	
	
	public static class Node{
		public int value;
		public Node(int v) {
			 value = v;
		}
	}
	
	public static void main(String[] args) {
		// UnSortedMap
		HashMap<Integer, String> map = new HashMap<>();
		map.put(1000000, "我是1000000");
		map.put(2, "我是2");
		map.put(3, "我是3");
		map.put(4, "我是4");
		map.put(5, "我是5");
		map.put(6, "我是6");
		map.put(1000000, "我是1000001");
		
		System.out.println(map.containsKey(1));
		System.out.println(map.containsKey(10));
		
		System.out.println(map.get(4));
		System.out.println(map.get(10));
		
		map.put(4, "他是4");
		System.out.println(map.get(4));
		
		map.remove(4);
		System.out.println(map.get(4));
		
		
		
		//       key
		HashSet<String>  set = new HashSet<>();
		set.add("abc");
		set.contains("abc");
		set.remove("abc");
		
		// 哈希表,增、删、改、查,在使用时,O(1)
		
		
		System.out.println("=====================");
		
		
		int a = 100000;
		int b = 100000;
		System.out.println(a == b);
		
		Integer c = 100000;
		Integer d = 100000;
		System.out.println(c.equals(d));
		
		Integer e = 127;  //  - 128  ~  127
		Integer f = 127;
		System.out.println(e == f);
		
		
		
		HashMap<Node, String> map2 = new HashMap<>();
		Node node1 = new Node(1);
		Node node2 = node1;
		map2.put(node1, "我是node1");
		map2.put(node2, "我是node1");
		System.out.println(map2.size());
		
		System.out.println("======================");
		
		TreeMap<Integer, String> treeMap = new TreeMap<>();
		
		treeMap.put(3, "我是3");
		treeMap.put(4, "我是4");
		treeMap.put(8, "我是8");
		treeMap.put(5, "我是5");
		treeMap.put(7, "我是7");
		treeMap.put(1, "我是1");
		treeMap.put(2, "我是2");

		System.out.println(treeMap.containsKey(1));
		System.out.println(treeMap.containsKey(10));
		
		System.out.println(treeMap.get(4));
		System.out.println(treeMap.get(10));
		
		treeMap.put(4, "他是4");
		System.out.println(treeMap.get(4));
		
		treeMap.remove(4);
		System.out.println(treeMap.get(4));
		
		System.out.println(treeMap.firstKey());
		System.out.println(treeMap.lastKey());
		// <= 4
		System.out.println(treeMap.floorKey(4));
		// >= 4
		System.out.println(treeMap.ceilingKey(4));
		
		// O(logN)	
		
	}
	
	
	
}
posted @ 2020-07-12 23:54  -Inky  阅读(172)  评论(1编辑  收藏