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《算法笔记》2. 链表、栈、队列、递归、哈希表、顺序表

目录

  • 链表、栈、队列、递归、哈希
    • 链表
      • 单向链表
      • 双向链表
      • 单双链表简单练习
    • 栈、队列
    • 栈、队列常见面试题
    • 递归
      • 递归行为的时间复杂度
    • 哈希表HashMap、HashSet
    • 顺序表 TreeMap、TreeSet

链表、栈、队列、递归、哈希

链表

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;
        }

    }

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;
        }
    }

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

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;
        }

    }

}

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!");

    }

}

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

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)

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

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)  

    }

}

文章永久链接:https://tech.souyunku.com/18687

未经允许不得转载:搜云库技术团队 » 《算法笔记》2. 链表、栈、队列、递归、哈希表、顺序表

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