Leetcode 题解 - 栈和队列

Leetcode 题解 - 栈和队列

• Leetcode 题解 - 栈和队列
  • 1. 用栈实现队列
  • 2. 用队列实现栈
  • 3. 最小值栈
  • 4. 用栈实现括号匹配
  • 5. 数组中元素与下一个比它大的元素之间的距离
  • 6. 循环数组中比当前元素大的下一个元素

1. 用栈实现队列

232. Implement Queue using Stacks (Easy)

Leetcode / 力扣

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();}
}

2. 用队列实现栈

225. Implement Stack using Queues (Easy)

Leetcode / 力扣

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();}
}

3. 最小值栈

155. Min Stack (Easy)

Leetcode / 力扣

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();}
}

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。

4. 用栈实现括号匹配

20. Valid Parentheses (Easy)

Leetcode / 力扣

"()[]{}"Output : true

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();
}

5. 数组中元素与下一个比它大的元素之间的距离

739. Daily Temperatures (Medium)

Leetcode / 力扣

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

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;
}

6. 循环数组中比当前元素大的下一个元素

503. Next Greater Element II (Medium)

Leetcode / 力扣

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.

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;
}