线性表 - 栈和队列

数组和链表都是线性存储结构的基础,栈和队列都是线性存储结构的应用。@pdai

知识点

栈 - LIFO

示意图

实现

  • 使用数组实现的叫静态栈
  • 使用链表实现的叫动态栈

队列 - FIFO

示意图

实现

  • 使用数组实现的叫静态队列
  • 使用链表实现的叫动态队列

JDK中实现

《Java - Stack & Queue 源码解析》

栈和队列相关题目

用栈实现队列

232. Implement Queue using Stacks (Easy) (opens new window)

栈的顺序为后进先出,而队列的顺序为先进先出。使用两个栈实现队列,一个元素需要经过两个栈才能出队列,在经过第一个栈时元素顺序被反转,经过第二个栈时再次被反转,此时就是先进先出顺序。

class MyQueue {

    private Stack<Integer> in = new Stack<>();
    private Stack<Integer> out = new Stack<>();

    public void push(int x) {
        in.push(x);
    }

    public int pop() {
        in2out();
        return out.pop();
    }

    public int peek() {
        in2out();
        return out.peek();
    }

    private void in2out() {
        if (out.isEmpty()) {
            while (!in.isEmpty()) {
                out.push(in.pop());
            }
        }
    }

    public boolean empty() {
        return in.isEmpty() && out.isEmpty();
    }
}
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

用队列实现栈

225. Implement Stack using Queues (Easy) (opens new window)

在将一个元素 x 插入队列时,为了维护原来的后进先出顺序,需要让 x 插入队列首部。而队列的默认插入顺序是队列尾部,因此在将 x 插入队列尾部之后,需要让除了 x 之外的所有元素出队列,再入队列。

class MyStack {

    private Queue<Integer> queue;

    public MyStack() {
        queue = new LinkedList<>();
    }

    public void push(int x) {
        queue.add(x);
        int cnt = queue.size();
        while (cnt-- > 1) {
            queue.add(queue.poll());
        }
    }

    public int pop() {
        return queue.remove();
    }

    public int top() {
        return queue.peek();
    }

    public boolean empty() {
        return queue.isEmpty();
    }
}
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

最小值栈

155. Min Stack (Easy) (opens new window)

class MinStack {

    private Stack<Integer> dataStack;
    private Stack<Integer> minStack;
    private int min;

    public MinStack() {
        dataStack = new Stack<>();
        minStack = new Stack<>();
        min = Integer.MAX_VALUE;
    }

    public void push(int x) {
        dataStack.add(x);
        min = Math.min(min, x);
        minStack.add(min);
    }

    public void pop() {
        dataStack.pop();
        minStack.pop();
        min = minStack.isEmpty() ? Integer.MAX_VALUE : minStack.peek();
    }

    public int top() {
        return dataStack.peek();
    }

    public int getMin() {
        return minStack.peek();
    }
}
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

对于实现最小值队列问题,可以先将队列使用栈来实现,然后就将问题转换为最小值栈,这个问题出现在 编程之美: 3.7。

用栈实现括号匹配

20. Valid Parentheses (Easy) (opens new window)

"()[]{}"

Output : true
1
2
3
public boolean isValid(String s) {
    Stack<Character> stack = new Stack<>();
    for (char c : s.toCharArray()) {
        if (c == '(' || c == '{' || c == '[') {
            stack.push(c);
        } else {
            if (stack.isEmpty()) {
                return false;
            }
            char cStack = stack.pop();
            boolean b1 = c == ')' && cStack != '(';
            boolean b2 = c == ']' && cStack != '[';
            boolean b3 = c == '}' && cStack != '{';
            if (b1 || b2 || b3) {
                return false;
            }
        }
    }
    return stack.isEmpty();
}
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20

数组中元素与下一个比它大的元素之间的距离

739. Daily Temperatures (Medium) (opens new window)

Input: [73, 74, 75, 71, 69, 72, 76, 73]
Output: [1, 1, 4, 2, 1, 1, 0, 0]
1
2

在遍历数组时用栈把数组中的数存起来,如果当前遍历的数比栈顶元素来的大,说明栈顶元素的下一个比它大的数就是当前元素。

public int[] dailyTemperatures(int[] temperatures) {
    int n = temperatures.length;
    int[] dist = new int[n];
    Stack<Integer> indexs = new Stack<>();
    for (int curIndex = 0; curIndex < n; curIndex++) {
        while (!indexs.isEmpty() && temperatures[curIndex] > temperatures[indexs.peek()]) {
            int preIndex = indexs.pop();
            dist[preIndex] = curIndex - preIndex;
        }
        indexs.add(curIndex);
    }
    return dist;
}
1
2
3
4
5
6
7
8
9
10
11
12
13

循环数组中比当前元素大的下一个元素

503. Next Greater Element II (Medium) (opens new window)

Input: [1,2,1]
Output: [2,-1,2]
Explanation: The first 1's next greater number is 2;
The number 2 can't find next greater number;
The second 1's next greater number needs to search circularly, which is also 2.
1
2
3
4
5

与 739. Daily Temperatures (Medium) 不同的是,数组是循环数组,并且最后要求的不是距离而是下一个元素。

public int[] nextGreaterElements(int[] nums) {
    int n = nums.length;
    int[] next = new int[n];
    Arrays.fill(next, -1);
    Stack<Integer> pre = new Stack<>();
    for (int i = 0; i < n * 2; i++) {
        int num = nums[i % n];
        while (!pre.isEmpty() && nums[pre.peek()] < num) {
            next[pre.pop()] = num;
        }
        if (i < n){
            pre.push(i);
        }
    }
    return next;
}
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16

参考文章

  • https://www.cnblogs.com/QG-whz/p/5170418.html

  • https://www.cnblogs.com/QG-whz/p/5171123.html

联系我

添加@pdai微信

PS:添加时请备注Java全栈,谢谢!