232 lines
6.7 KiB
Java
232 lines
6.7 KiB
Java
<!-- GFM-TOC -->
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* [1. 用栈实现队列](#1-用栈实现队列)
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* [2. 用队列实现栈](#2-用队列实现栈)
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* [3. 最小值栈](#3-最小值栈)
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* [4. 用栈实现括号匹配](#4-用栈实现括号匹配)
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* [5. 数组中元素与下一个比它大的元素之间的距离](#5-数组中元素与下一个比它大的元素之间的距离)
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* [6. 循环数组中比当前元素大的下一个元素](#6-循环数组中比当前元素大的下一个元素)
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<!-- GFM-TOC -->
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# 1. 用栈实现队列
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[232. Implement Queue using Stacks (Easy)](https://leetcode.com/problems/implement-queue-using-stacks/description/)
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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。
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```java
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class MyQueue {
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private Stack<Integer> in = new Stack<>();
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private Stack<Integer> out = new Stack<>();
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public void push(int x) {
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in.push(x);
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}
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public int pop() {
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in2out();
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return out.pop();
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}
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public int peek() {
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in2out();
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return out.peek();
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}
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private void in2out() {
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if (out.isEmpty()) {
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while (!in.isEmpty()) {
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out.push(in.pop());
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}
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}
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}
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public boolean empty() {
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return in.isEmpty() && out.isEmpty();
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}
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}
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```
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# 2. 用队列实现栈
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[225. Implement Stack using Queues (Easy)](https://leetcode.com/problems/implement-stack-using-queues/description/)
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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。
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```java
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class MyStack {
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private Queue<Integer> queue;
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public MyStack() {
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queue = new LinkedList<>();
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}
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public void push(int x) {
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queue.add(x);
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int cnt = queue.size();
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while (cnt-- > 1) {
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queue.add(queue.poll());
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}
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}
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public int pop() {
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return queue.remove();
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}
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public int top() {
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return queue.peek();
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}
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public boolean empty() {
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return queue.isEmpty();
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}
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}
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```
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# 3. 最小值栈
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[155. Min Stack (Easy)](https://leetcode.com/problems/min-stack/description/)
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```java
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class MinStack {
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private Stack<Integer> dataStack;
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private Stack<Integer> minStack;
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private int min;
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public MinStack() {
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dataStack = new Stack<>();
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minStack = new Stack<>();
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min = Integer.MAX_VALUE;
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}
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public void push(int x) {
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dataStack.add(x);
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min = Math.min(min, x);
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minStack.add(min);
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}
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public void pop() {
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dataStack.pop();
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minStack.pop();
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min = minStack.isEmpty() ? Integer.MAX_VALUE : minStack.peek();
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}
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public int top() {
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return dataStack.peek();
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}
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public int getMin() {
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return minStack.peek();
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}
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}
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```
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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。
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# 4. 用栈实现括号匹配
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[20. Valid Parentheses (Easy)](https://leetcode.com/problems/valid-parentheses/description/)
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```html
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"()[]{}"
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Output : true
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```
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```java
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public boolean isValid(String s) {
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Stack<Character> stack = new Stack<>();
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for (char c : s.toCharArray()) {
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if (c == '(' || c == '{' || c == '[') {
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stack.push(c);
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} else {
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if (stack.isEmpty()) {
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return false;
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}
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char cStack = stack.pop();
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boolean b1 = c == ')' && cStack != '(';
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boolean b2 = c == ']' && cStack != '[';
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boolean b3 = c == '}' && cStack != '{';
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if (b1 || b2 || b3) {
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return false;
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}
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}
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}
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return stack.isEmpty();
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}
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```
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# 5. 数组中元素与下一个比它大的元素之间的距离
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[739. Daily Temperatures (Medium)](https://leetcode.com/problems/daily-temperatures/description/)
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```html
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Input: [73, 74, 75, 71, 69, 72, 76, 73]
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Output: [1, 1, 4, 2, 1, 1, 0, 0]
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```
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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。
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```java
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public int[] dailyTemperatures(int[] temperatures) {
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int n = temperatures.length;
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int[] dist = new int[n];
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Stack<Integer> indexs = new Stack<>();
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for (int curIndex = 0; curIndex < n; curIndex++) {
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while (!indexs.isEmpty() && temperatures[curIndex] > temperatures[indexs.peek()]) {
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int preIndex = indexs.pop();
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dist[preIndex] = curIndex - preIndex;
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}
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indexs.add(curIndex);
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}
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return dist;
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}
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```
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# 6. 循环数组中比当前元素大的下一个元素
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[503. Next Greater Element II (Medium)](https://leetcode.com/problems/next-greater-element-ii/description/)
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```text
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Input: [1,2,1]
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Output: [2,-1,2]
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Explanation: The first 1's next greater number is 2;
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The number 2 can't find next greater number;
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The second 1's next greater number needs to search circularly, which is also 2.
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```
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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。
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```java
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public int[] nextGreaterElements(int[] nums) {
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int n = nums.length;
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int[] next = new int[n];
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Arrays.fill(next, -1);
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Stack<Integer> pre = new Stack<>();
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for (int i = 0; i < n * 2; i++) {
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int num = nums[i % n];
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while (!pre.isEmpty() && nums[pre.peek()] < num) {
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next[pre.pop()] = num;
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}
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if (i < n){
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pre.push(i);
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}
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}
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return next;
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}
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```
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# 微信公众号
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更多精彩内容将发布在微信公众号 CyC2018 上，你也可以在公众号后台和我交流学习和求职相关的问题。另外，公众号提供了该项目的 PDF 等离线阅读版本，后台回复 "下载" 即可领取。公众号也提供了一份技术面试复习大纲，不仅系统整理了面试知识点，而且标注了各个知识点的重要程度，从而帮你理清多而杂的面试知识点，后台回复 "大纲" 即可领取。我基本是按照这个大纲来进行复习的，对我拿到了 BAT 头条等 Offer 起到很大的帮助。你们完全可以和我一样根据大纲上列的知识点来进行复习，就不用看很多不重要的内容，也可以知道哪些内容很重要从而多安排一些复习时间。
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