227 lines
6.0 KiB
Markdown
227 lines
6.0 KiB
Markdown
<!-- GFM-TOC -->
|
||
* [用栈实现队列](#用栈实现队列)
|
||
* [用队列实现栈](#用队列实现栈)
|
||
* [最小值栈](#最小值栈)
|
||
* [用栈实现括号匹配](#用栈实现括号匹配)
|
||
* [数组中元素与下一个比它大的元素之间的距离](#数组中元素与下一个比它大的元素之间的距离)
|
||
* [循环数组中比当前元素大的下一个元素](#循环数组中比当前元素大的下一个元素)
|
||
<!-- GFM-TOC -->
|
||
|
||
|
||
# 用栈实现队列
|
||
|
||
[232. Implement Queue using Stacks (Easy)](https://leetcode.com/problems/implement-queue-using-stacks/description/)
|
||
|
||
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。
|
||
|
||
```java
|
||
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();
|
||
}
|
||
}
|
||
```
|
||
|
||
# 用队列实现栈
|
||
|
||
[225. Implement Stack using Queues (Easy)](https://leetcode.com/problems/implement-stack-using-queues/description/)
|
||
|
||
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。
|
||
|
||
```java
|
||
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();
|
||
}
|
||
}
|
||
```
|
||
|
||
# 最小值栈
|
||
|
||
[155. Min Stack (Easy)](https://leetcode.com/problems/min-stack/description/)
|
||
|
||
```java
|
||
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。
|
||
|
||
# 用栈实现括号匹配
|
||
|
||
[20. Valid Parentheses (Easy)](https://leetcode.com/problems/valid-parentheses/description/)
|
||
|
||
```html
|
||
"()[]{}"
|
||
|
||
Output : true
|
||
```
|
||
|
||
```java
|
||
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();
|
||
}
|
||
```
|
||
|
||
# 数组中元素与下一个比它大的元素之间的距离
|
||
|
||
[739. Daily Temperatures (Medium)](https://leetcode.com/problems/daily-temperatures/description/)
|
||
|
||
```html
|
||
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。
|
||
|
||
```java
|
||
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;
|
||
}
|
||
```
|
||
|
||
# 循环数组中比当前元素大的下一个元素
|
||
|
||
[503. Next Greater Element II (Medium)](https://leetcode.com/problems/next-greater-element-ii/description/)
|
||
|
||
```text
|
||
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。
|
||
|
||
```java
|
||
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;
|
||
}
|
||
```
|
||
|
||
|
||
|
||
|
||
</br><div align="center">欢迎关注公众号，获取最新文章！</div></br></br>
|
||
<div align="center"><img width="180px" src="https://cyc-1256109796.cos.ap-guangzhou.myqcloud.com/%E5%85%AC%E4%BC%97%E5%8F%B7.jpg"></img></div>
|