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数据结构二:栈+队列+递归(DataWhale系列)-云栖社区-阿里云

2019-05-15

Datawhale 系列数据结构 Task2 1 栈 2 1 1用数组实现一个顺序栈 public class ArrayStack { private T [] data; private int size; private int cnt; @

Datawhale 系列数据结构 Task2.1 栈 2.1.1用数组实现一个顺序栈 public class ArrayStack { private T [] data; private int size; private int cnt; @...

Datawhale 系列数据结构

Task2.1 栈

2.1.1用数组实现一个顺序栈

public class ArrayStack<T> {
    private T [] data;
    private int size;
    private int cnt;
    
    @SuppressWarnings("unchecked")
    public ArrayStack(){
        data= (T[]) new Object [10]; 
        cnt = 0;
        size =10;
    }
    
    public void push(T t){
        if(cnt>=size){
            data=Arrays.copyOf(data, data.length*2);
            size*=2;
        }
        data[size] = t;
        cnt++;
    }
    public T peek(){
        if (cnt==0) {
            throw new EmptyStackException();
        }
        return data[cnt];
    }
    public T pop(){
        if (cnt==0) {
            throw new EmptyStackException();
        }
        cnt--;
        return data[cnt];
    }
    public int search(T t){
        for(int i=cnt;i>0;i--){
            if(data[i]==t)
                return i;
        }
        return -1;
    }
    public boolean isEmpty(){
        return cnt==0;
    }
}

2.1.2 用链表实现一个链式栈

public class ListStack<T> {
    private List<T> data;
    private int cnt;
    
    public ListStack(){
        data= new LinkedList<T>(); 
        cnt = 0;
    }
    
    public void push(T t){
        data.add(t);
        cnt++;
    }
    public T peek(){
        if (cnt==0) {
            throw new EmptyStackException();
        }
        return data.get(cnt);
    }
    public T pop(){
        if (cnt==0) {
            throw new EmptyStackException();
        }
        T t=data.remove(cnt);
        cnt--;
        return t; 
    }
    public int search(T t){
        for(int i=cnt;i>0;i--){
            if(data.get(i)==t)
                return i;
        }
        return -1;
    }
    public boolean isEmpty(){
        return cnt==0;
    }
}

2.1.3 编程模拟实现一个浏览器的前进后退功能

public class BrowserSimula {
    public static void main(String[] args) {
        Browser b1=new Browser();
        b1.open();
        b1.open();
        b1.open();
        b1.open();
        System.out.println(b1.backward());
        System.out.println(b1.backward());
        System.out.println(b1.forward());
        System.out.println(b1.forward());
    }
}

class Browser{
    private Stack<Integer> s1;
    private Stack<Integer> s2;
    int cnt;
    
    Browser(){
        s1=new Stack<Integer>();
        s2=new Stack<Integer>();
        cnt=0;
    }
    /**
     * 点开一个新的链接
     */
    public void open(){
        cnt++;
        s1.push(cnt);
    }
    //后退
    public Integer backward(){
        if(cnt==0)
            throw new ArrayIndexOutOfBoundsException(cnt);
        Integer a =s1.pop();
        s2.push(a);
        return s1.peek();
    }
    //前进
    public Integer forward(){
        if(s2.isEmpty())
            throw new ArrayIndexOutOfBoundsException(cnt);
        Integer a =s2.pop();
        s1.push(a);
        return a;
    }
    
}

2.1.4 练习

//Valid Parentheses(有效的括号)
 private HashMap<Character, Character> mappings;

  public Solution() {
    this.mappings = new HashMap<Character, Character>();
    this.mappings.put(')', '(');
    this.mappings.put('}', '{');
    this.mappings.put(']', '[');
  }

  public boolean isValid(String s) {
    Stack<Character> stack = new Stack<Character>();

    for (int i = 0; i < s.length(); i++) {
      char c = s.charAt(i);
      if (this.mappings.containsKey(c)) {
        char topElement = stack.empty() ? '#' : stack.pop();
        if (topElement != this.mappings.get(c)) {
          return false;
        }
      } else {
        stack.push(c);
      }
    }
    return stack.isEmpty();
  }
//Longest Valid Parentheses(最长有效的括号)[作为可选]
/*dp 方法:
我们用 dp[i] 表示以 i 结尾的最长有效括号;

1 当 s[i] 为 ( , dp[i] 必然等于0,因为不可能组成有效的括号;
2 那么 s[i] 为 )
    2.1 当 s[i-1] 为 (,那么 dp[i] = dp[i-2] + 2;
    2.2 当 s[i-1] 为 ) 并且 s[i-dp[i-1] - 1] 为 (,那么 dp[i] = dp[i-1] + 2 + dp[i-dp[i-1]-2];
时间复杂度:O(n)
*/
public int longestValidParentheses(String s) {
    if (s == null || s.length() == 0) return 0;
    int[] dp = new int[s.length()];
    int res = 0;
    for (int i = 0; i < s.length(); i++) {
        if (i > 0 && s.charAt(i) == ')') {
            if (s.charAt(i - 1) == '(') {
                dp[i] = (i - 2 >= 0 ? dp[i - 2] + 2 : 2);
            } else if (s.charAt(i - 1) == ')' && i - dp[i - 1] - 1 >= 0 && s.charAt(i - dp[i - 1] - 1) == '(') {
                dp[i] = dp[i - 1] + 2 + (i - dp[i - 1] - 2 >= 0 ? dp[i - dp[i - 1] - 2] : 0);
             }
        }
    res = Math.max(res, dp[i]);
    }
    return res;
}

//Evaluate Reverse Polish Notatio(逆波兰表达式求值)
 public int evalRPN(String[] tokens) {
        Stack<Integer> stack = new Stack<>();
        for (int i = 0 ;i < tokens.length;i++){
            String str = tokens[i];
            if (str.length() == 1){
                char ch = str.charAt(0);
                if (ch-'0' >= 0 && ch - '0' <= 9 ){
                    Integer a = Integer.valueOf(str);
                    stack.push(a);
                }                    
                else{
                    if (stack.size() < 2)
                        return 0;
                    int num2 = stack.pop();
                    int num1 = stack.pop();
                    switch(ch){
                        case '+':                            
                            stack.push(num1 + num2);
                            break;
                        case '-':
                            stack.push(num1 - num2);
                            break;
                        case '*':
                            stack.push(num1 * num2);
                            break;
                        case '/':
                            stack.push(num1 / num2);
                            break;
                    }
                    
                }
            }else{
                int n = Integer.valueOf(str);
                stack.push(n);
            }
        }
        return stack.pop(); 
    }

TASK2.2 队列

2.2.1 用数组实现一个顺序队列

public class ArrayQueue<T>  {
    private T [] data ;
    private int cnt;//元素个数    
    private int size;//队列长度
    
    @SuppressWarnings("unchecked")
    public ArrayQueue(){
        data =(T[]) new Object [10];
        cnt = 0;
        size = 10;
    }
    @SuppressWarnings("unchecked")
    public ArrayQueue(int size){
        data =(T[]) new Object [size];
        cnt = 0;
        this.size = size;
    }
    
    public void add(T t){
        if(cnt>=size){
            throw new IllegalStateException();
        }
        data[cnt]=t;
        cnt++;
    }
    
    public T remove(){
        if(cnt<0){
            throw new NoSuchElementException();
        }
        T t= data[0];
        data = Arrays.copyOfRange(data,0,size);
        cnt--;
        return t;
        
    }    
    public boolean offer(T t){
        if(cnt>=size){
            return false;
        }
        data[cnt]=t;
        cnt++;
        return true;
    }    
    public boolean pull(T t){
        if(cnt<0){
            return false;
        }
        data = Arrays.copyOfRange(data,0,size);
        cnt--;
        return true;
    }
    //返回队列头元素
    public T element(){
        return data[0];    
    }     
    public boolean isEmpty(){
        return cnt==0 ;
    }
    public boolean isFull(){
        return cnt==size;
    }
}

2.2.2 用链表实现一个链式队列

public class ListQueue<T> {
    private List<T> data ;
    private int cnt;//元素个数
    
    private int size;//队列长度(用链表的话这里可以强行定义)
    
    public ListQueue(){
        data =new LinkedList<T>();
        cnt = 0;
        size = 10;
    }

    public void add(T t){
        data.add(t);
        cnt++;
    }
    
    public T remove(){
        if(cnt<0){
            throw new NoSuchElementException();
        }
        T t= data.remove(cnt);
        cnt--;
        return t;
        
    }
    
    public boolean offer(T t){
        if(cnt>=size){
            return false;
        }
        data.add(t);
        cnt++;
        return true;
    }
    
    public boolean pull(T t){
        if(cnt<0){
            return false;
        }
        data.remove(cnt);
        cnt--;
        return true;
    }
    //返回队列头元素
    public T element(){
        return data.get(0);    
    }
    
    public boolean isEmpty(){
        return cnt==0 ;
    }
    public boolean isFull(){
        return cnt==size;
    }
}

2.2.3 实现一个循环队列

public class MyCircularQueue {
    private final int capacity;
    private final int[] array;
    private int head = 0;
    private int tail = 0;
    private int count = 0;

    /**
     * Initialize your data structure here. Set the size of the queue to be k.
     */
    public MyCircularQueue(int k) {
        this.capacity = k;
        this.array = new int[this.capacity];
    }

    /**
     * Insert an element into the circular queue. Return true if the operation is successful.
     */
    public boolean enQueue(int value) {
        //队列已满
        if (count == capacity) {
            return false;
        }

        //队列为空, 重新设置头部
        if (count == 0) {
            head = (head == capacity) ? 0 : head;
            tail = head;
            array[head] = value;
            count++;
            return true;
        }

        //队列未满 (有空位)
        if (tail == capacity - 1) {
            //tail 达到 maxIndex, 重置为 0
            tail = 0;
        } else {
            //tail 未达到 maxIndex, tail++
            tail++;
        }
        array[tail] = value;
        count++;
        return true;
    }

    /**
     * Delete an element from the circular queue. Return true if the operation is successful.
     */
    public boolean deQueue() {
        if (count == 0) {
            //队列为空
            return false;
        }
        count--;
        head++;
        return true;
    }

    /**
     * Get the front item from the queue.
     */
    public int Front() {
        if (count == 0 ) {
            return -1;
        }
        return array[head];
    }

    /**
     * Get the last item from the queue.
     */
    public int Rear() {
        if (count == 0 ) {
            return -1;
        }
        return array[tail];
    }

    /**
     * Checks whether the circular queue is empty or not.
     */
    public boolean isEmpty() {
        return count == 0;
    }

    /**
     * Checks whether the circular queue is full or not.
     */
    public boolean isFull() {
        return count == capacity;
    }
}

2.2.4 练习

//Design Circular Deque(设计一个双端队列)
class MyCircularDeque {
    public int k;
    public int[] numbers;
    public int head;
    public int tail;
    /** Initialize your data structure here. Set the size of the deque to be k. */
    public MyCircularDeque(int k) {
        numbers=new int[k+1];
        head=0;
        tail=0;
        this.k=k;
    }
    
    /** Adds an item at the front of Deque. Return true if the operation is successful. */
    public boolean insertFront(int value) {
        if(isFull())
            return false;
        else{
            head=(head+k)%(k+1);
            numbers[head]=value;
            return true;
        }
    }
    
    /** Adds an item at the rear of Deque. Return true if the operation is successful. */
    public boolean insertLast(int value) {
        if(isFull())
            return false;
        else{
            numbers[tail]=value;
            tail=(tail+1)%(k+1);
            return true;
        }
    }
    
    /** Deletes an item from the front of Deque. Return true if the operation is successful. */
    public boolean deleteFront() {
         if(isEmpty())
            return false;
        else{
            head=(head+1)%(k+1);
            return true;
        }
    }
    
    /** Deletes an item from the rear of Deque. Return true if the operation is successful. */
    public boolean deleteLast() {
         if(isEmpty())
            return false;
        else{
            tail=(tail+k)%(k+1);
            return true;
        }
    }
    
    /** Get the front item from the deque. */
    public int getFront() {
        if(isEmpty())
            return -1;
        return numbers[head];
    }
    
    /** Get the last item from the deque. */
    public int getRear() {
        if(isEmpty())
            return -1;
        return numbers[(tail+k)%(k+1)];
    }
    
    /** Checks whether the circular deque is empty or not. */
    public boolean isEmpty() {
        return tail==head;
    }
    
    /** Checks whether the circular deque is full or not. */
    public boolean isFull() {
        return (tail+1)%(k+1)==head;
    }
}
//Sliding Window Maximum(滑动窗口最大值)
class Solution {
    public int[] maxSlidingWindow(int[] nums, int k) {
        if(k==0){
            return new int[0];
        }
        List<Integer> list = new ArrayList<>();//存放窗口内的数字
        int max = Integer.MIN_VALUE;//窗口内的最大数字
        for(int i = 0; i<k;i++){
            if(max<nums[i]){
                max = nums[i];
            }
            list.add(nums[i]);
        }
        int[] res = new int[nums.length - k + 1];//要返回的结果数据
        res[0] = max;
        for(int i = k; i < nums.length;i++){
            int z =list.remove(0);//移走第一位数并插入新的一位数
            list.add(nums[i]);
            if(z!=max){//移走的数不是max,只需判断max与新插入的数哪个大
                if(nums[i]> max){
                    max = nums[i];
                }
                res[i-k+1] = max;
            }else{//移走的数是max,重新判断列表中哪个数是最大的
                if(!list.contains(max)){
                    max = Integer.MIN_VALUE;
                    for(Integer num : list){
                        if(max<num){
                            max = num;
                        }
                    }
                }else{
                    if(nums[i]> max){
                        max = nums[i];
                    }
                }    
            }
            res[i-k+1] = max;
        }
        return res;
    }
}

3 递归

3.1 编程实现斐波那契数列求值 f(n)=f(n-1)+f(n-2)

public static int Fibe(int n){
        if(n==1) return 1;
        if(n==2) return 1;
        return Fibe(n-1)+Fibe(n-2);
    }

3.2 编程实现求阶乘 n!

public static int factorial(int n){
        if(n==1) return 1;
        return factorial(n-1)*n;
    }

3.3 编程实现一组数据集合的全排列

class Solution {
    List<List<Integer>> res = new ArrayList<>();
    
    public List<List<Integer>> permute(int[] nums) {
        permition(nums, 0, nums.length-1);
        return res;
    }
    
    public void permition(int[] nums, int p, int q){
        if(p==q){
            res.add(arrayToList(nums));
        }
        for(int i = p; i <= q; i++){
            swap(nums, i, p);
            permition(nums, p+1, q);
            swap(nums, i,p);
        }
    }
    
    private List<Integer> arrayToList(int[] nums){
        List<Integer> res = new ArrayList<>();
        for(int i = 0; i < nums.length; i++){
            res.add(nums[i]);
        }
        
        return res;
    }
    
    private void swap(int[] nums, int i, int j){
        int temp = nums[i];
        nums[i] = nums[j];
        nums[j] = temp;
    }
}

3.4 练习

// Climbing Stairs(爬楼梯)
//直接递归,但是直接递归leetcode竟然不能通过,于是又方法2
class Solution {
    public int climbStairs(int n) {
        if(n==1) return 1;
        if(n==2) return 2;
        return  climbStairs(n-1)+climbStairs(n-2);
    }
}

//非递归放啊
class Solution {
    public int climbStairs(int n) {
        int [] ways = new int[n+1];
        ways[0] = 0;
        for (int i = 1;i<ways.length;i++){
            if (i < 3 ){
                ways[i] = i;
            }else {
                ways[i] = ways[i-1] + ways[i-2];
            }
        }
        return ways[n];
    }
}
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