算法初级面试题03——队列实现栈、栈实现队列、转圈打印矩阵、旋转矩阵、反转链表、之字打印矩阵、排序矩阵中找数

时间：2019-01-16 19:16:06 阅读：128 评论：0 收藏：0 [点我收藏+]

标签：LLC ati bec hint lca you 调整 close null

第一部分主要讨论：栈、队列、数组矩阵相关的面试题

题目一

用数组结构实现大小固定的队列和栈

    public static class ArrayStack {
        private Integer[] arr;
        private Integer size;

        public ArrayStack(int initSize) {
            if (initSize < 0) {
                throw new IllegalArgumentException("The init size is less than 0");
            }
            arr = new Integer[initSize];
            size = 0;
        }

        public Integer peek() {
            if (size == 0) {
                return null;
            }
            return arr[size - 1];
        }

        public void push(int obj) {
            if (size == arr.length) {
                throw new ArrayIndexOutOfBoundsException("The queue is full");
            }
            arr[size++] = obj;
        }

        public Integer pop() {
            if (size == 0) {
                throw new ArrayIndexOutOfBoundsException("The queue is empty");
            }
            return arr[--size];
        }
    }

    public static class ArrayQueue {
        private Integer[] arr;
        private Integer size;
        private Integer first;
        private Integer last;

        public ArrayQueue(int initSize) {
            if (initSize < 0) {
                throw new IllegalArgumentException("The init size is less than 0");
            }
            arr = new Integer[initSize];
            size = 0;
            first = 0;
            last = 0;
        }

        public Integer peek() {
            if (size == 0) {
                return null;
            }
            return arr[first];
        }

        public void push(int obj) {
            if (size == arr.length) {
                throw new ArrayIndexOutOfBoundsException("The queue is full");
            }
            size++;
            arr[last] = obj;
            last = last == arr.length - 1 ? 0 : last + 1;
        }

        public Integer poll() {
            if (size == 0) {
                throw new ArrayIndexOutOfBoundsException("The queue is empty");
            }
            size--;
            int tmp = first;
            first = first == arr.length - 1 ? 0 : first + 1;
            return arr[tmp];
        }
    }

实现代码

题目二

实现一个特殊的栈，在实现栈的基本功能的基础上，再实现返回栈中最小元素的操作。

【要求】

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：利用两个栈来实现

public class Code_02_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) {
            if (this.stackMin.isEmpty()) {
                this.stackMin.push(newNum);
            } else if (newNum <= this.getmin()) {
                this.stackMin.push(newNum);
            }
            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.");
            }
            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());
    }

}

实现代码

技术分享图片

题目三

如何仅用队列结构实现栈结构？

思路：两个队列复制到只剩下一个，留着最晚进入的然后给用户。

技术分享图片

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

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

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

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

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

实现代码

如何仅用栈结构实现队列结构？

思路：①push要一次倒完 ②如果pop有东西一定不要倒

技术分享图片

    public static class TwoQueuesStack {
        private Queue<Integer> queue;
        private Queue<Integer> help;

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

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

        public int peek() {
            if (queue.isEmpty()) {
                throw new RuntimeException("Stack is empty!");
            }
            while (queue.size() != 1) {
                help.add(queue.poll());
            }
            int res = queue.poll();
            help.add(res);
            swap();
            return res;
        }

        public int pop() {
            if (queue.isEmpty()) {
                throw new RuntimeException("Stack is empty!");
            }
            while (queue.size() > 1) {
                help.add(queue.poll());
            }
            int res = queue.poll();
            swap();
            return res;
        }

        private void swap() {
            Queue<Integer> tmp = help;
            help = queue;
            queue = tmp;
        }

    }

实现代码

题目四

猫狗队列【题目】宠物、狗和猫的类如下：

public class Pet {

　　private String type; public Pet(String type) { this.type = type;

}

public String getPetType() { return this.type; } }

public class Dog extends Pet { public Dog() { super("dog"); } }

public class Cat extends Pet { public Cat() { super("cat"); } }

实现一种狗猫队列的结构，要求如下：

用户可以调用add方法将cat类或dog类的实例放入队列中；

用户可以调用pollAll方法，将队列中所有的实例按照进队列的先后顺序依次弹出；

用户可以调用pollDog方法，将队列中dog类的实例按照进队列的先后顺序依次弹出；

用户可以调用pollCat方法，将队列中cat类的实例按照进队列的先后顺序依次弹出；

用户可以调用isEmpty方法，检查队列中是否还有dog或cat的实例；

用户可以调用isDogEmpty方法，检查队列中是否有dog类的实例；

用户可以调用isCatEmpty方法，检查队列中是否有cat类的实例。

（思路很简单就加多一个时间戳count变量来区分，哪个先进入）

public class Code_04_DogCatQueue {

    public static class Pet {
        private String type;

        public Pet(String type) {
            this.type = type;
        }

        public String getPetType() {
            return this.type;
        }
    }

    public static class Dog extends Pet {
        public Dog() {
            super("dog");
        }
    }

    public static class Cat extends Pet {
        public Cat() {
            super("cat");
        }
    }

    public static class PetEnterQueue {
        private Pet pet;
        private long count;

        public PetEnterQueue(Pet pet, long count) {
            this.pet = pet;
            this.count = count;
        }

        public Pet getPet() {
            return this.pet;
        }

        public long getCount() {
            return this.count;
        }

        public String getEnterPetType() {
            return this.pet.getPetType();
        }
    }

    public static class DogCatQueue {
        private Queue<PetEnterQueue> dogQ;
        private Queue<PetEnterQueue> catQ;
        private long count;

        public DogCatQueue() {
            this.dogQ = new LinkedList<PetEnterQueue>();
            this.catQ = new LinkedList<PetEnterQueue>();
            this.count = 0;
        }

        public void add(Pet pet) {
            if (pet.getPetType().equals("dog")) {
                this.dogQ.add(new PetEnterQueue(pet, this.count++));
            } else if (pet.getPetType().equals("cat")) {
                this.catQ.add(new PetEnterQueue(pet, this.count++));
            } else {
                throw new RuntimeException("err, not dog or cat");
            }
        }

        public Pet pollAll() {
            if (!this.dogQ.isEmpty() && !this.catQ.isEmpty()) {
                if (this.dogQ.peek().getCount() < this.catQ.peek().getCount()) {
                    return this.dogQ.poll().getPet();
                } else {
                    return this.catQ.poll().getPet();
                }
            } else if (!this.dogQ.isEmpty()) {
                return this.dogQ.poll().getPet();
            } else if (!this.catQ.isEmpty()) {
                return this.catQ.poll().getPet();
            } else {
                throw new RuntimeException("err, queue is empty!");
            }
        }

        public Dog pollDog() {
            if (!this.isDogQueueEmpty()) {
                return (Dog) this.dogQ.poll().getPet();
            } else {
                throw new RuntimeException("Dog queue is empty!");
            }
        }

        public Cat pollCat() {
            if (!this.isCatQueueEmpty()) {
                return (Cat) this.catQ.poll().getPet();
            } else
                throw new RuntimeException("Cat queue is empty!");
        }

        public boolean isEmpty() {
            return this.dogQ.isEmpty() && this.catQ.isEmpty();
        }

        public boolean isDogQueueEmpty() {
            return this.dogQ.isEmpty();
        }

        public boolean isCatQueueEmpty() {
            return this.catQ.isEmpty();
        }

    }

    public static void main(String[] args) {
        DogCatQueue test = new DogCatQueue();

        Pet dog1 = new Dog();
        Pet cat1 = new Cat();
        Pet dog2 = new Dog();
        Pet cat2 = new Cat();
        Pet dog3 = new Dog();
        Pet cat3 = new Cat();

        test.add(dog1);
        test.add(cat1);
        test.add(dog2);
        test.add(cat2);
        test.add(dog3);
        test.add(cat3);

        test.add(dog1);
        test.add(cat1);
        test.add(dog2);
        test.add(cat2);
        test.add(dog3);
        test.add(cat3);

        test.add(dog1);
        test.add(cat1);
        test.add(dog2);
        test.add(cat2);
        test.add(dog3);
        test.add(cat3);
        while (!test.isDogQueueEmpty()) {
            System.out.println(test.pollDog().getPetType());
        }
        while (!test.isEmpty()) {
            System.out.println(test.pollAll().getPetType());
        }
    }

}

实现代码

题目五

锻炼宏观思路解题

转圈打印矩阵

【题目】给定一个整型矩阵matrix，请按照转圈的方式打印它。

打印结果为：1，2，3，4，8，12，16，15，14，13，9，5，6，7，11， 10

【要求】额外空间复杂度为O(1)。

思路：

技术分享图片

package class_03;

public class Code_06_PrintMatrixSpiralOrder {

    public static void spiralOrderPrint(int[][] matrix) {
        //左上角
        int tR = 0;
        int tC = 0;
        //右下角
        int dR = matrix.length - 1;//行数
        int dC = matrix[0].length - 1;//列数
        while (tR <= dR && tC <= dC) {
            //左上角和右下角
            printEdge(matrix, tR++, tC++, dR--, dC--);
        }
    }

    public static void printEdge(int[][] m, int tR, int tC, int dR, int dC) {
        if (tR == dR) {//行先相遇，打印目标是横的长方形
            for (int i = tC; i <= dC; i++) {
                System.out.print(m[tR][i] + " ");
            }
        } else if (tC == dC) {//列先相遇，打印目标是竖的长方形
            for (int i = tR; i <= dR; i++) {
                System.out.print(m[i][tC] + " ");
            }
        } else {
            int curC = tC;
            int curR = tR;
            //模拟转圈打印
            while (curC != dC) {
                System.out.print(m[tR][curC] + " ");
                curC++;
            }
            while (curR != dR) {
                System.out.print(m[curR][dC] + " ");
                curR++;
            }
            while (curC != tC) {
                System.out.print(m[dR][curC] + " ");
                curC--;
            }
            while (curR != tR) {
                System.out.print(m[curR][tC] + " ");
                curR--;
            }
        }
    }

    public static void main(String[] args) {
        int[][] matrix = { { 1, 2, 3, 4 }, { 5, 6, 7, 8 }, { 9, 10, 11, 12 },
                { 13, 14, 15, 16 } };
        spiralOrderPrint(matrix);

    }

}

实现代码

题目六

旋转正方形矩阵

【题目】给定一个整型正方形矩阵matrix，请把该矩阵调整成顺时针旋转90度的样子。

【要求】额外空间复杂度为O(1)。

思路：

技术分享图片

package class_03;

public class Code_05_RotateMatrix {

    public static void rotate(int[][] matrix) {
        int tR = 0;
        int tC = 0;
        int dR = matrix.length - 1;
        int dC = matrix[0].length - 1;
        while (tR < dR) {
            rotateEdge(matrix, tR++, tC++, dR--, dC--);
        }
    }

    public static void rotateEdge(int[][] m, int ax, int ay, int bx, int by) {
        int times = by - ay;
        int tmp = 0;
        for (int i = 0; i != times; i++) {
            tmp = m[ax][ay + i];
            m[ax][ay + i] = m[bx - i][ay];
            m[bx - i][ay] = m[bx][by - i];
            m[bx][by - i] = m[ax + i][by];
            m[ax + i][by] = tmp;
        }
    }

    public static void printMatrix(int[][] matrix) {
        for (int i = 0; i != matrix.length; i++) {
            for (int j = 0; j != matrix[0].length; j++) {
                System.out.print(matrix[i][j] + " ");
            }
            System.out.println();
        }
    }

    public static void main(String[] args) {
        int[][] matrix = { { 1, 2, 3, 4 }, { 5, 6, 7, 8 }, { 9, 10, 11, 12 },
                { 13, 14, 15, 16 } };
        printMatrix(matrix);
        rotate(matrix);
        System.out.println("=========");
        printMatrix(matrix);

    }

}

实现代码

题目七

反转单向和双向链表

【题目】分别实现反转单向链表和反转双向链表的函数。

【要求】如果链表长度为N，时间复杂度要求为O(N)，额外空间复杂度要求为O(1)

package class_03;

public class Code_07_ReverseList {

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

    public static Node reverseList(Node head) {
        Node pre = null;
        Node next = null;
        while (head != null) {
            //把下一个节点先存起来
            next = head.next;
            //翻转节点指向
            head.next = pre;
            //把当前节点设置为下一个节点的前节点
            pre = head;
            //为下一次循环，推进一步
            head = next;
        }
        return pre;
    }

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

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

    public static DoubleNode reverseList(DoubleNode head) {
        DoubleNode pre = null;
        DoubleNode next = null;
        while (head != null) {
            next = head.next;
            head.next = pre;
            head.last = next;
            //把当前节点设置为下一个节点的前节点
            pre = head;
            //为下一循环做准备，往下一个要操作的节点移动
            head = next;
        }
        return pre;
    }

    public static void printLinkedList(Node head) {
        System.out.print("Linked List: ");
        while (head != null) {
            System.out.print(head.value + " ");
            head = head.next;
        }
        System.out.println();
    }

    public static void printDoubleLinkedList(DoubleNode head) {
        System.out.print("Double Linked List: ");
        DoubleNode end = null;
        while (head != null) {
            System.out.print(head.value + " ");
            end = head;
            head = head.next;
        }
        System.out.print("| ");
        while (end != null) {
            System.out.print(end.value + " ");
            end = end.last;
        }
        System.out.println();
    }

    public static void main(String[] args) {
        Node head1 = new Node(1);
        head1.next = new Node(2);
        head1.next.next = new Node(3);
        printLinkedList(head1);
        head1 = reverseList(head1);
        printLinkedList(head1);

        DoubleNode head2 = new DoubleNode(1);
        head2.next = new DoubleNode(2);
        head2.next.last = head2;
        head2.next.next = new DoubleNode(3);
        head2.next.next.last = head2.next;
        head2.next.next.next = new DoubleNode(4);
        head2.next.next.next.last = head2.next.next;
        printDoubleLinkedList(head2);
        printDoubleLinkedList(reverseList(head2));

    }

}

实现代码

题目八

“之”字形打印矩阵

【题目】给定一个矩阵matrix，按照“之”字形的方式打印这个矩阵，例如： 1 2 3 4 5 6 7 8 9 10 11 12 “之”字形打印的结果为：1，2，5，9，6，3，4，7，10，11，8，12

技术分享图片

【要求】额外空间复杂度为O(1)。

思路：

技术分享图片

package class_03;

public class Code_08_ZigZagPrintMatrix {

    public static void printMatrixZigZag(int[][] matrix) {
        int ax = 0;
        int ay = 0;
        int bx = 0;
        int by = 0;
        int endX = matrix.length - 1;
        int endY = matrix[0].length - 1;
        boolean fromUp = false;
        while (ax != endX + 1) {
            printLevel(matrix, ax, ay, bx, by, fromUp);
            ax = ay == endY ? ax + 1 : ax;
            ay = ay == endY ? ay : ay + 1;
            by = bx == endX ? by + 1 : by;
            bx = bx == endX ? bx : bx + 1;
            fromUp = !fromUp;
        }
        System.out.println();
    }

    public static void printLevel(int[][] m, int ax, int ay, int bx, int by,
            boolean f) {
        if (f) {
            while (ax != bx + 1) {
                System.out.print(m[ax++][ay--] + " ");
            }
        } else {
            while (bx != ax - 1) {
                System.out.print(m[bx--][by++] + " ");
            }
        }
    }

    public static void main(String[] args) {
        int[][] matrix = { { 1, 2, 3, 4 }, { 5, 6, 7, 8 }, { 9, 10, 11, 12 } };
        printMatrixZigZag(matrix);

    }

}

实现代码

题目九

在行列都排好序的矩阵中找数

【题目】给定一个有N*M的整型矩阵matrix和一个整数K，matrix的每一行和每一列都是排好序的。实现一个函数，判断K是否在matrix中。

例如： 0 1 2 5 2 3 4 7 4 4 4 8 5 7 7 9

如果K为7，返回true；如果K为6，返回false。

【要求】时间复杂度为O(N+M)，额外空间复杂度为O(1)。

思路：

技术分享图片

package class_03;

public class Code_09_FindNumInSortedMatrix {

    public static boolean isContains(int[][] matrix, int K) {
        int row = 0;
        int col = matrix[0].length - 1;
        while (row < matrix.length && col > -1) {
            if (matrix[row][col] == K) {
                return true;
            } else if (matrix[row][col] > K) {
                col--;
            } else {
                row++;
            }
        }
        return false;
    }

    public static void main(String[] args) {
        int[][] matrix = new int[][] { { 0, 1, 2, 3, 4, 5, 6 },// 0
                { 10, 12, 13, 15, 16, 17, 18 },// 1
                { 23, 24, 25, 26, 27, 28, 29 },// 2
                { 44, 45, 46, 47, 48, 49, 50 },// 3
                { 65, 66, 67, 68, 69, 70, 71 },// 4
                { 96, 97, 98, 99, 100, 111, 122 },// 5
                { 166, 176, 186, 187, 190, 195, 200 },// 6
                { 233, 243, 321, 341, 356, 370, 380 } // 7
        };
        int K = 233;
        System.out.println(isContains(matrix, K));
    }

}

实现代码

算法初级面试题03——队列实现栈、栈实现队列、转圈打印矩阵、旋转矩阵、反转链表、之字打印矩阵、排序矩阵中找数

标签：LLC ati bec hint lca you 调整 close null

原文地址：https://www.cnblogs.com/xieyupeng/p/10278468.html

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