Haiyang Yu's Blog
0%

Leetcode 225 Implement Stack Using Queues

发表于 更新于 分类于 Valine：
本文字数： 5.2k 阅读时长 ≈ 5 分钟

Implement a last in first out (LIFO) stack using only two queues. The implemented stack should support all the functions of a normal queue (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 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.

Follow-up: Can you implement the stack 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.

Example 1:

1
2
3
4
5
6
7
8
9
10
11
12
13
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.

1 2个队列. push O(1), pop O(n)

2个队列互相交替, 插入的时候维持原有顺序, pop的时候把含有element的队列的前n-1个元素转移到另外一个队列, 第n个元素pop出来.

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
class MyStack {
    Queue<Integer> a;
    Queue<Integer> b;
    /** Initialize your data structure here. */
    public MyStack() {
        a = new ArrayDeque<>();
        b = new ArrayDeque<>();
    }
    
    /** Push element x onto stack. */
    public void push(int x) {
        Queue<Integer> full;
        Queue<Integer> empty;
        if(!a.isEmpty())
        {
            full = a;
            empty = b;
        }
        else
        {
            full = b;
            empty = a;
        }
        full.offer(x);
    }
    
    /** Removes the element on top of the stack and returns that element. */
    public int pop() {
        Queue<Integer> full;
        Queue<Integer> empty;
        if(!a.isEmpty())
        {
            full = a;
            empty = b;
        }
        else
        {
            full = b;
            empty = a;
        }
        while(full.size() > 1)
        {
            int tmp = full.poll();
            empty.offer(tmp);
        }
        return full.poll();
    }
    
    /** Get the top element. */
    public int top() {
        Queue<Integer> full;
        Queue<Integer> empty;
        if(!a.isEmpty())
        {
            full = a;
            empty = b;
        }
        else
        {
            full = b;
            empty = a;
        }
        while(full.size() > 1)
        {
            int tmp = full.poll();
            empty.offer(tmp);
        }
        int ans = full.peek();
        empty.offer(full.poll());
        return ans;
    }
    
    /** Returns whether the stack is empty. */
    public boolean empty() {
        return a.isEmpty() && b.isEmpty();
    }
}

/**
 * Your MyStack object will be instantiated and called as such:
 * MyStack obj = new MyStack();
 * obj.push(x);
 * int param_2 = obj.pop();
 * int param_3 = obj.top();
 * boolean param_4 = obj.empty();
 */

2 2个队列. push O(n), pop O(1)

插入的时候就把顺序导致过来, 通过两个队列的交替始终保持最新插入的元素在队列第一个位置. pop和top只需要O(1)的操作

感觉这一种方法比第一种好一些.

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
class MyStack {
    Queue<Integer> a;
    Queue<Integer> b;
    /** Initialize your data structure here. */
    public MyStack() {
        a = new ArrayDeque<>();
        b = new ArrayDeque<>();
    }
    
    /** Push element x onto stack. */
    public void push(int x) {
        Queue<Integer> full;
        Queue<Integer> empty;
        if(!a.isEmpty())
        {
            full = a;
            empty = b;
        }
        else
        {
            full = b;
            empty = a;
        }
        empty.offer(x);
        while(!full.isEmpty())
        {
            empty.offer(full.poll());
        }
    }
    
    /** Removes the element on top of the stack and returns that element. */
    public int pop() {
        Queue<Integer> full;
        Queue<Integer> empty;
        if(!a.isEmpty())
        {
            full = a;
            empty = b;
        }
        else
        {
            full = b;
            empty = a;
        }
        return full.poll();
    }
    
    /** Get the top element. */
    public int top() {
        Queue<Integer> full;
        Queue<Integer> empty;
        if(!a.isEmpty())
        {
            full = a;
            empty = b;
        }
        else
        {
            full = b;
            empty = a;
        }
        return full.peek();
    }
    
    /** Returns whether the stack is empty. */
    public boolean empty() {
        return a.isEmpty() && b.isEmpty();
    }
}

/**
 * Your MyStack object will be instantiated and called as such:
 * MyStack obj = new MyStack();
 * obj.push(x);
 * int param_2 = obj.pop();
 * int param_3 = obj.top();
 * boolean param_4 = obj.empty();
 */

3 1个队列

这种方法真是妙啊.

还是插入的时候就要颠倒次序.

假如已经插入了1,2,3,4.

这时队列为 4,3,2,1.

当插入5的时候, 先记录队列的大小size.

然后插入5, 队列变为4,3,2,1,5

然后队列首元素出队列再进队列size次

即4,3,2,1,5 -> 3,2,1,5,4 -> 2,1,5,4,3 -> 1,5,4,3,2 -> 5,4,3,2,1

这样就完成了队列到栈的转换.

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
class MyStack {
    Queue<Integer> a;
    /** Initialize your data structure here. */
    public MyStack() {
        a = new ArrayDeque<>();
    }
    
    /** Push element x onto stack. */
    public void push(int x) {
        int size = a.size();
        a.offer(x);
        while(size > 0)
        {
            a.offer(a.poll());
            --size;
        }
    }
    
    /** Removes the element on top of the stack and returns that element. */
    public int pop() {
        return a.poll();
    }
    
    /** Get the top element. */
    public int top() {
        return a.peek();
    }
    
    /** Returns whether the stack is empty. */
    public boolean empty() {
        return a.isEmpty();
    }
}

/**
 * Your MyStack object will be instantiated and called as such:
 * MyStack obj = new MyStack();
 * obj.push(x);
 * int param_2 = obj.pop();
 * int param_3 = obj.top();
 * boolean param_4 = obj.empty();
 */