代码随想录训练营｜Day 11｜Stack & Queue, 232, 225, 20, 1047

标签：11 20 int 随想录 element queue push return public

Stack & Queue

队列是先进先出
栈是先进后出

QUEUE

在 FIFO 数据结构中，将首先处理添加到队列中的第一个元素。
插入（insert）操作也称作入队（enqueue），新元素始终被添加在队列的末尾。删除（delete）操作也被称为出队（dequeue)。你只能移除第一个元素。

Implementation:

// "static void main" must be defined in a public class.

class MyQueue {
    // store elements
    private List<Integer> data;         
    // a pointer to indicate the start position
    private int p_start;            
    public MyQueue() {
        data = new ArrayList<Integer>();
        p_start = 0;
    }
    /** Insert an element into the queue. Return true if the operation is successful. */
    public boolean enQueue(int x) {
        data.add(x);
        return true;
    };    
    /** Delete an element from the queue. Return true if the operation is successful. */
    public boolean deQueue() {
        if (isEmpty() == true) {
            return false;
        }
        p_start++;
        return true;
    }
    /** Get the front item from the queue. */
    public int Front() {
        return data.get(p_start);
    }
    /** Checks whether the queue is empty or not. */
    public boolean isEmpty() {
        return p_start >= data.size();
    }     
};

public class Main {
    public static void main(String[] args) {
        MyQueue q = new MyQueue();
        q.enQueue(5);
        q.enQueue(3);
        if (q.isEmpty() == false) {
            System.out.println(q.Front());
        }
        q.deQueue();
        if (q.isEmpty() == false) {
            System.out.println(q.Front());
        }
        q.deQueue();
        if (q.isEmpty() == false) {
            System.out.println(q.Front());
        }
    }
}

可以使用固定大小的数组和两个指针来指示起始位置和结束位置。目的是重用我们之前提到的被浪费的存储。
在循环队列中，我们使用一个数组和两个指针（head 和 tail）。 head 表示队列的起始位置，tail 表示队列的结束位置。

Implementation:

class MyCircularQueue {
    
    private int[] data;
    private int head;
    private int tail;
    private int size;

    /** Initialize your data structure here. Set the size of the queue to be k. */
    public MyCircularQueue(int k) {
        data = new int[k];
        head = -1;
        tail = -1;
        size = k;
    }
    
    /** Insert an element into the circular queue. Return true if the operation is successful. */
    public boolean enQueue(int value) {
        if (isFull() == true) {
            return false;
        }
        if (isEmpty() == true) {
            head = 0;
        }
        tail = (tail + 1) % size;
        data[tail] = value;
        return true;
    }
    
    /** Delete an element from the circular queue. Return true if the operation is successful. */
    public boolean deQueue() {
        if (isEmpty() == true) {
            return false;
        }
        if (head == tail) {
            head = -1;
            tail = -1;
            return true;
        }
        head = (head + 1) % size;
        return true;
    }
    
    /** Get the front item from the queue. */
    public int Front() {
        if (isEmpty() == true) {
            return -1;
        }
        return data[head];
    }
    
    /** Get the last item from the queue. */
    public int Rear() {
        if (isEmpty() == true) {
            return -1;
        }
        return data[tail];
    }
    
    /** Checks whether the circular queue is empty or not. */
    public boolean isEmpty() {
        return head == -1;
    }
    
    /** Checks whether the circular queue is full or not. */
    public boolean isFull() {
        return ((tail + 1) % size) == head;
    }
}

/**
 * Your MyCircularQueue object will be instantiated and called as such:
 * MyCircularQueue obj = new MyCircularQueue(k);
 * boolean param_1 = obj.enQueue(value);
 * boolean param_2 = obj.deQueue();
 * int param_3 = obj.Front();
 * int param_4 = obj.Rear();
 * boolean param_5 = obj.isEmpty();
 * boolean param_6 = obj.isFull();
 */

Examples:

// "static void main" must be defined in a public class.
public class Main {
    public static void main(String[] args) {
        // 1. Initialize a queue.
        Queue<Integer> q = new LinkedList();
        // 2. Get the first element - return null if queue is empty.
        System.out.println("The first element is: " + q.peek());
        // 3. Push new element.
        q.offer(5);
        q.offer(13);
        q.offer(8);
        q.offer(6);
        // 4. Pop an element.
        q.poll();
        // 5. Get the first element.
        System.out.println("The first element is: " + q.peek());
        // 7. Get the size of the queue.
        System.out.println("The size is: " + q.size());
    }
}

STACK

在 LIFO 数据结构中，将首先处理添加到队列中的最新元素。

与队列不同，栈是一个 LIFO 数据结构。通常，插入操作在栈中被称作入栈 push 。与队列类似，总是在堆栈的末尾添加一个新元素。但是，删除操作，退栈 pop ，将始终删除队列中相对于它的最后一个元素。

Implementation:

// "static void main" must be defined in a public class.
class MyStack {
    private List<Integer> data;               // store elements
    public MyStack() {
        data = new ArrayList<>();
    }
    /** Insert an element into the stack. */
    public void push(int x) {
        data.add(x);
    }
    /** Checks whether the queue is empty or not. */
    public boolean isEmpty() {
        return data.isEmpty();
    }
    /** Get the top item from the queue. */
    public int top() {
        return data.get(data.size() - 1);
    }
    /** Delete an element from the queue. Return true if the operation is successful. */
    public boolean pop() {
        if (isEmpty()) {
            return false;
        }
        data.remove(data.size() - 1);
        return true;
    }
};

public class Main {
    public static void main(String[] args) {
        MyStack s = new MyStack();
        s.push(1);
        s.push(2);
        s.push(3);
        for (int i = 0; i < 4; ++i) {
            if (!s.isEmpty()) {
                System.out.println(s.top());
            }
            System.out.println(s.pop());
        }
    }
}

Examples:

// "static void main" must be defined in a public class.
public class Main {
    public static void main(String[] args) {
        // 1. Initialize a stack.
        Stack<Integer> s = new Stack<>();
        // 2. Push new element.
        s.push(5);
        s.push(13);
        s.push(8);
        s.push(6);
        // 3. Check if stack is empty.
        if (s.empty() == true) {
            System.out.println("Stack is empty!");
            return;
        }
        // 4. Pop an element.
        s.pop();
        // 5. Get the top element.
        System.out.println("The top element is: " + s.peek());
        // 6. Get the size of the stack.
        System.out.println("The size is: " + s.size());
    }
}

For Future References

文章讲解： https://leetcode.cn/leetbook/read/queue-stack/xkrhpg/

232. Implement Queue using Stacks

Implement a first in first out (FIFO) queue using only two stacks. The implemented queue should support all the functions of a normal queue (push, peek, pop, and empty).

Implement the MyQueue class:

void push(int x) Pushes element x to the back of the queue.
int pop() Removes the element from the front of the queue and returns it.
int peek() Returns the element at the front of the queue.
boolean empty() Returns true if the queue is empty, false otherwise.

Notes:

You must use only standard operations of a stack, which means only push to top, peek/pop from top, size, and is empty operations are valid.
Depending on your language, the stack may not be supported natively. You may simulate a stack using a list or deque (double-ended queue) as long as you use only a stack's standard operations.

Example 1:

Input
["MyQueue", "push", "push", "peek", "pop", "empty"]
[[], [1], [2], [], [], []]
Output
[null, null, null, 1, 1, false]

Explanation
MyQueue myQueue = new MyQueue();
myQueue.push(1); // queue is: [1]
myQueue.push(2); // queue is: [1, 2] (leftmost is front of the queue)
myQueue.peek(); // return 1
myQueue.pop(); // return 1, queue is [2]
myQueue.empty(); // return false

Constraints:

1 <= x <= 9
At most 100 calls will be made to push, pop, peek, and empty.
All the calls to pop and peek are valid.

Follow-up: Can you implement the queue such that each operation is amortized O(1) time complexity? In other words, performing n operations will take overall O(n) time even if one of those operations may take longer.

需要两个栈一个输入栈，一个输出栈，这里要注意输入栈和输出栈的关系。
在push数据的时候，只要数据放进输入栈就好，但在pop的时候，操作就复杂一些，输出栈如果为空，就把进栈数据全部导入进来（注意是全部导入），再从出栈弹出数据，如果输出栈不为空，则直接从出栈弹出数据就可以了。
如果进栈和出栈都为空的话，说明模拟的队列为空了。

class MyQueue {

    Stack<Integer> stackIn;
    Stack<Integer> stackOut;

    /** Initialize your data structure here. */
    public MyQueue() {
        stackIn = new Stack<>(); // 负责进栈
        stackOut = new Stack<>(); // 负责出栈
    }
    
    /** Push element x to the back of queue. */
    public void push(int x) {
        stackIn.push(x);
    }
    
    /** Removes the element from in front of queue and returns that element. */
    public int pop() {    
        dumpstackIn();
        return stackOut.pop();
    }
    
    /** Get the front element. */
    public int peek() {
        dumpstackIn();
        return stackOut.peek();
    }
    
    /** Returns whether the queue is empty. */
    public boolean empty() {
        return stackIn.isEmpty() && stackOut.isEmpty();
    }

    // 如果stackOut为空，那么将stackIn中的元素全部放到stackOut中
    private void dumpstackIn(){
        if (!stackOut.isEmpty()) return; 
        while (!stackIn.isEmpty()){
                stackOut.push(stackIn.pop());
        }
    }
}

For Future References

题目链接：https://leetcode.com/problems/implement-queue-using-stacks/

文章讲解： https://programmercarl.com/0232.用栈实现队列.html

视频讲解：https://www.bilibili.com/video/BV1nY4y1w7VC/

225. Implement Stack using Queues

Implement a last-in-first-out (LIFO) stack using only two queues. The implemented stack should support all the functions of a normal stack (push, top, pop, and empty).

Implement the MyStack class:

void push(int x) Pushes element x to the top of the stack.
int pop() Removes the element on the top of the stack and returns it.
int top() Returns the element on the top of the stack.
boolean empty() Returns true if the stack is empty, false otherwise.

Notes:

You must use only standard operations of a queue, which means that only push to back, peek/pop from front, size and is empty operations are valid.
Depending on your language, the queue may not be supported natively. You may simulate a queue using a list or deque (double-ended queue) as long as you use only a queue's standard operations.

Example 1:

Input
["MyStack", "push", "push", "top", "pop", "empty"]
[[], [1], [2], [], [], []]
Output
[null, null, null, 2, 2, false]

Explanation
MyStack myStack = new MyStack();
myStack.push(1);
myStack.push(2);
myStack.top(); // return 2
myStack.pop(); // return 2
myStack.empty(); // return False

Constraints:

1 <= x <= 9
At most 100 calls will be made to push, pop, top, and empty.
All the calls to pop and top are valid.

Follow-up: Can you implement the stack using only one queue?

两个队列来实现栈的思路:

队列是先进先出的规则，把一个队列中的数据导入另一个队列中，数据的顺序并没有变，并没有变成先进后出的顺序。
用两个队列que1和que2实现队列的功能，que2其实完全就是一个备份的作用，把que1最后面的元素以外的元素都备份到que2，然后弹出最后面的元素，再把其他元素从que2导回que1。

class MyStack {

    Queue<Integer> queue1; // 和栈中保持一样元素的队列
    Queue<Integer> queue2; // 辅助队列

    /** Initialize your data structure here. */
    public MyStack() {
        queue1 = new LinkedList<>();
        queue2 = new LinkedList<>();
    }
    
    /** Push element x onto stack. */
    public void push(int x) {
        queue2.offer(x); // 先放在辅助队列中
        while (!queue1.isEmpty()){
            queue2.offer(queue1.poll());
        }
        Queue<Integer> queueTemp;
        queueTemp = queue1;
        queue1 = queue2;
        queue2 = queueTemp; // 最后交换queue1和queue2，将元素都放到queue1中
    }
    
    /** Removes the element on top of the stack and returns that element. */
    public int pop() {
        return queue1.poll(); // 因为queue1中的元素和栈中的保持一致，所以这个和下面两个的操作只看queue1即可
    }
    
    /** Get the top element. */
    public int top() {
        return queue1.peek();
    }
    
    /** Returns whether the stack is empty. */
    public boolean empty() {
        return queue1.isEmpty();
    }
}

其实这道题目就是用一个队列就够了。

一个队列在模拟栈弹出元素的时候只要将队列头部的元素（除了最后一个元素外）重新添加到队列尾部，此时在去弹出元素就是栈的顺序了。

class MyStack {
    // Deque 接口继承了 Queue 接口
    // 所以 Queue 中的 add、poll、peek等效于 Deque 中的 addLast、pollFirst、peekFirst
    Deque<Integer> que1;
    /** Initialize your data structure here. */
    public MyStack() {
        que1 = new ArrayDeque<>();
    }
    
    /** Push element x onto stack. */
    public void push(int x) {
        que1.addLast(x);
    }
    
    /** Removes the element on top of the stack and returns that element. */
    public int pop() {
        int size = que1.size();
        size--;
        // 将 que1 导入 que2 ，但留下最后一个值
        while (size-- > 0) {
            que1.addLast(que1.peekFirst());
            que1.pollFirst();
        }

        int res = que1.pollFirst();
        return res;
    }
    
    /** Get the top element. */
    public int top() {
        return que1.peekLast();
    }
    
    /** Returns whether the stack is empty. */
    public boolean empty() {
        return que1.isEmpty();
    }
}

For Future References

题目链接：https://leetcode.com/problems/implement-stack-using-queues/

文章讲解： https://programmercarl.com/0225.用队列实现栈.html

视频讲解：https://www.bilibili.com/video/BV1Fd4y1K7sm/

20. Valid Parentheses

Given a string s containing just the characters '(', ')', '{', '}', '[' and ']', determine if the input string is valid.

An input string is valid if:

Open brackets must be closed by the same type of brackets.
Open brackets must be closed in the correct order.
Every close bracket has a corresponding open bracket of the same type.

Example 1:

Input: s = "()"
Output: true

Example 2:

Input: s = "()[]{}"
Output: true

Example 3:

Input: s = "(]"
Output: false

Constraints:

1 <= s.length <= 104
s consists of parentheses only '()[]{}'.

括号匹配是使用栈解决的经典问题。

题意其实就像我们在写代码的过程中，要求括号的顺序是一样的，有左括号，相应的位置必须要有右括号。

编译器在词法分析的过程中处理括号、花括号等这个符号的逻辑，也是使用了栈这种数据结构。

有三种不匹配的情况，

字符串里左方向的括号多余了，所以不匹配。
第二种情况，括号没有多余，但是括号的类型没有匹配上。
第三种情况，字符串里右方向的括号多余了，所以不匹配。

第一种情况：已经遍历完了字符串，但是栈不为空，说明有相应的左括号没有右括号来匹配，所以return false

第二种情况：遍历字符串匹配的过程中，发现栈里没有要匹配的字符。所以return false

第三种情况：遍历字符串匹配的过程中，栈已经为空了，没有匹配的字符了，说明右括号没有找到对应的左括号return false

那么什么时候说明左括号和右括号全都匹配了呢，就是字符串遍历完之后，栈是空的，就说明全都匹配了。

class Solution {
    public boolean isValid(String s) {
        Deque<Character> deque = new LinkedList<>();
        char ch;
        for (int i = 0; i < s.length(); i++) {
            ch = s.charAt(i);
            //碰到左括号，就把相应的右括号入栈
            if (ch == '(') {
                deque.push(')');
            }else if (ch == '{') {
                deque.push('}');
            }else if (ch == '[') {
                deque.push(']');
            } else if (deque.isEmpty() || deque.peek() != ch) {
                return false;
            }else {//如果是右括号判断是否和栈顶元素匹配
                deque.pop();
            }
        }
        //最后判断栈中元素是否匹配
        return deque.isEmpty();
    }
}

Time Complexity：O(n)
Space Complexity：O(n)

For Future References

题目链接：https://leetcode.com/problems/valid-parentheses/

文章讲解： https://programmercarl.com/0020.有效的括号.html

视频讲解：https://www.bilibili.com/video/BV1AF411w78g/

1047. Remove All Adjacent Duplicates In String

You are given a string s consisting of lowercase English letters. A duplicate removal consists of choosing two adjacent and equal letters and removing them.

We repeatedly make duplicate removals on s until we no longer can.

Return the final string after all such duplicate removals have been made. It can be proven that the answer is unique.

Example 1:

Input: s = "abbaca"
Output: "ca"
Explanation:
For example, in "abbaca" we could remove "bb" since the letters are adjacent and equal, and this is the only possible move.  The result of this move is that the string is "aaca", of which only "aa" is possible, so the final string is "ca".

Example 2:

Input: s = "azxxzy"
Output: "ay"

Constraints:

1 <= s.length <= 105
s consists of lowercase English letters.

栈的目的，就是存放遍历过的元素，当遍历当前的这个元素的时候，去栈里看一下我们是不是遍历过相同数值的相邻元素。
因为从栈里弹出的元素是倒序的，所以在对字符串进行反转一下，就得到了最终的结果。

Deque

class Solution {
    public String removeDuplicates(String S) {
        //ArrayDeque会比LinkedList在除了删除元素这一点外会快一点
        //参考：https://stackoverflow.com/questions/6163166/why-is-arraydeque-better-than-linkedlist
        ArrayDeque<Character> deque = new ArrayDeque<>();
        char ch;
        for (int i = 0; i < S.length(); i++) {
            ch = S.charAt(i);
            if (deque.isEmpty() || deque.peek() != ch) {
                deque.push(ch);
            } else {
                deque.pop();
            }
        }
        String str = "";
        //剩余的元素即为不重复的元素
        while (!deque.isEmpty()) {
            str = deque.pop() + str;
        }
        return str;
    }
}

拿字符串直接作为栈

class Solution {
    public String removeDuplicates(String s) {
        // 将 res 当做栈
        StringBuffer res = new StringBuffer();
        // top为 res 的长度
        int top = -1;
        for (int i = 0; i < s.length(); i++) {
            char c = s.charAt(i);
            // 当 top > 0,即栈中有字符时，当前字符如果和栈中字符相等，弹出栈顶字符，同时 top--
            if (top >= 0 && res.charAt(top) == c) {
                res.deleteCharAt(top);
                top--;
            // 否则，将该字符 入栈，同时top++
            } else {
                res.append(c);
                top++;
            }
        }
        return res.toString();
    }
}

Two Pointers

class Solution {
    public String removeDuplicates(String s) {
        char[] ch = s.toCharArray();
        int fast = 0;
        int slow = 0;
        while(fast < s.length()){
            // 直接用fast指针覆盖slow指针的值
            ch[slow] = ch[fast];
            // 遇到前后相同值的，就跳过，即slow指针后退一步，下次循环就可以直接被覆盖掉了
            if(slow > 0 && ch[slow] == ch[slow - 1]){
                slow--;
            }else{
                slow++;
            }
            fast++;
        }
        return new String(ch,0,slow);
    }
}

Time Complexity：O(n)
Space Complexity：O(1)

For Future References

题目链接：https://leetcode.com/problems/remove-all-adjacent-duplicates-in-string/

文章讲解： https://programmercarl.com/1047.删除字符串中的所有相邻重复项.html

视频讲解：https://www.bilibili.com/video/BV12a411P7mw/

标签：11,20,int,随想录,element,queue,push,return,public
From： https://www.cnblogs.com/bluesociety/p/16746674.html

代码随想录训练营｜Day 11｜Stack & Queue, 232, 225, 20, 1047

Stack & Queue

232. Implement Queue using Stacks

225. Implement Stack using Queues

20. Valid Parentheses

1047. Remove All Adjacent Duplicates In String

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