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  • LC.225. Implement Stack using Queues(using two queues)

    https://leetcode.com/problems/implement-stack-using-queues/description/
    Implement the following operations of a stack using queues.

    push(x) -- Push element x onto stack.
    pop() -- Removes the element on top of the stack.
    top() -- Get the top element.
    empty() -- Return whether the stack is empty.
    Notes:
    You must use only standard operations of a queue -- which means only push to back, peek/pop from front, size, and is empty operations are valid.
    Depending on your language, queue may not be supported natively. You may simulate a queue by using a list or deque (double-ended queue), as long as you use only standard operations of a queue.
    You may assume that all operations are valid (for example, no pop or top operations will be called on an empty stack).


    第一个做法:复杂,啰嗦,受到 impelement queue using two stacks 的影响 queue 和 stack 不一样,倒腾多次并不会发生结构的变化
     1 public class LC_225_ImplementStackUsingTwoQueues_I {
     2 
     3     private Queue<Integer> queue1;
     4     private Queue<Integer> queue2;
     5 
     6     public LC_225_ImplementStackUsingTwoQueues_I() {
     7         queue1 = new LinkedList<>();
     8         queue2 = new LinkedList<>();
     9     }
    10 
    11     /**
    12      * Push element x onto stack.
    13      */
    14     public void push(int x) {
    15         if (!queue1.isEmpty()) {
    16             queue1.offer(x);
    17         } else {
    18             queue2.offer(x);
    19         }
    20     }
    21 
    22     /**
    23      * Removes the element on top of the stack and returns that element.
    24      */
    25     public int pop() {
    26         if (!queue1.isEmpty() && queue2.isEmpty()) {
    27             int size = queue1.size();
    28             shuffle(queue1, queue2, size - 1);
    29             return queue1.poll();
    30         } else {
    31             int size = queue2.size();
    32             shuffle(queue2, queue1, size - 1);
    33             return queue2.poll();
    34         }
    35     }
    36 
    37     /**
    38      * Get the top element.: peek
    39      */
    40     public int top() {
    41         int value = 0;
    42         if (!queue1.isEmpty() && queue2.isEmpty()) {
    43             int size = queue1.size();
    44             shuffle(queue1, queue2, size - 1);
    45             value = queue1.poll();
    46             queue2.offer(value);
    47         } else {
    48             int size = queue2.size();
    49             shuffle(queue2, queue1, size - 1);
    50             value = queue2.poll();
    51             queue1.offer(value);
    52         }
    53         return value;
    54     }
    55 
    56     //make sure queue1 and queue2 only one has values
    57     private void shuffle(Queue<Integer> from, Queue<Integer> to, int times) {
    58         for (int i = 0; i < times; i++) {
    59             to.offer(from.poll());
    60         }
    61     }
    62 
    63     /**
    64      * Returns whether the stack is empty.
    65      */
    66     public boolean empty() {
    67         return queue1.isEmpty() && queue2.isEmpty();
    68     }
    69 }

    方法2: 换指针,queue1 永远都是 非空, 倒腾到 queue2 中后,改变 reference 就行。注意这里 cut reference 的好习惯

     1 public class LC_225_ImplementStackUsingTwoQueues_II {
     2 
     3     private Queue<Integer> queue1;
     4     private Queue<Integer> queue2;
     5 
     6     public LC_225_ImplementStackUsingTwoQueues_II() {
     7         queue1 = new LinkedList<>();
     8         queue2 = new LinkedList<>();
     9     }
    10 
    11     /**
    12      * Push element x onto stack.
    13      */
    14     public void push(int x) {
    15         queue1.offer(x);
    16     }
    17 
    18     /**
    19      * Removes the element on top of the stack and returns that element.
    20      */
    21     public int pop() {
    22         int size = queue1.size();
    23         shuffle(queue1, queue2, size - 1);
    24         int value =  queue1.poll();
    25         //cut the reference
    26         queue1 = new LinkedList<>(queue2) ;
    27         queue2.clear();
    28         return value ;
    29     }
    30 
    31     /**
    32      * Get the top element.: peek
    33      */
    34     public int top() {
    35         int value = 0;
    36         int size = queue1.size();
    37         shuffle(queue1, queue2, size - 1);
    38         value = queue1.poll();
    39         queue2.offer(value);
    40         //cut the reference
    41         queue1 = new LinkedList<>(queue2) ;
    42         queue2.clear();
    43         return value;
    44     }
    45 
    46     //make sure queue1 and queue2 only one has values
    47     private void shuffle(Queue<Integer> from, Queue<Integer> to, int times) {
    48         for (int i = 0; i < times; i++) {
    49             to.offer(from.poll());
    50         }
    51     }
    52 
    53     /**
    54      * Returns whether the stack is empty.
    55      */
    56     public boolean empty() {
    57         return queue1.isEmpty() && queue2.isEmpty();
    58     }
    59 }
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  • 原文地址:https://www.cnblogs.com/davidnyc/p/8598796.html
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