有效的括号
解法：压栈出栈
public class Solution {
public boolean isValid(String s) {
Deque<Character> deque = new LinkedList<>();
Map<Character, Character> map = new HashMap<>();
map.put(')', '(');
map.put(']', '[');
map.put('}', '{');
for (char c : s.toCharArray()) {
if (deque.isEmpty() || deque.peekFirst() != map.get(c)) {
deque.offerFirst(c);
} else {
deque.pollFirst();
}
}
return deque.isEmpty();
}
public static void main(String[] args) {
boolean valid = new Solution().isValid("()[]][");
System.out.println(valid);
}
}
最小栈
思路：
- 运用两个栈，维护一个值栈和一个最小值栈
- 入栈时判断值是否大于最小值栈栈顶，如果大于等于就复制一个栈顶元素到栈顶
- 如果入栈元素小于最小值栈顶，加入最小值栈
class MinStack {
Deque<Integer> deque;
Deque<Integer> minDeque;
/** * initialize your data structure here. */
public MinStack() {
//值栈
deque = new LinkedList<>();
//最小值栈
minDeque = new LinkedList<>();
}
public void push(int x) {
//入值栈
deque.offerFirst(x);
//如果x小于最小值栈顶，就如最小值栈
if (minDeque.isEmpty() || minDeque.peekFirst() >= x) {
minDeque.offerFirst(x);
} else {
//如果最小值栈数小于x，就拷贝一份最小值在栈顶
minDeque.offerFirst(minDeque.peekFirst());
}
}
public void pop() {
deque.pollFirst();
minDeque.pollFirst();
}
public int top() {
return deque.peekFirst();
}
public int getMin() {
return minDeque.peekFirst();
}
}
用栈实现队列
class MyQueue {
Stack<Integer> in;
Stack<Integer> out;
/** * Initialize your data structure here. */
public MyQueue() {
in = new Stack<>();
out = new Stack<>();
}
/** * Push element x to the back of queue. */
public void push(int x) {
in.push(x);
}
/** * Removes the element from in front of queue and returns that element. */
public int pop() {
if (out.isEmpty()) {
while (!in.isEmpty()) {
out.push(in.pop());
}
}
return out.pop();
}
/** * Get the front element. */
public int peek() {
if (out.isEmpty()) {
while (!in.isEmpty()) {
out.push(in.pop());
}
}
return out.peek();
}
/** * Returns whether the queue is empty. */
public boolean empty() {
return in.isEmpty() && out.isEmpty();
}
}
用队列实现栈
思路：维护两个队列，保证总有一个队列是空的
- push操作，加入到非空的队列中
- pop操作，将队列中size-1的数加入到空的队列，留下最后一个pop出去
- top操作，将队列中size-1的数加入到空的队列，留下最后一个记录为top，再加入到新队列，返回top
- empty操作，两个队列同时满足为空
class MyStack {
LinkedList<Integer> queue1;
LinkedList<Integer> queue2;
/** * Initialize your data structure here. */
public MyStack() {
queue1 = new LinkedList<>();
queue2 = new LinkedList<>();
}
/** * Push element x onto stack. */
public void push(int x) {
if (queue1.isEmpty()) {
queue2.offer(x);
} else {
queue1.offer(x);
}
}
/** * Removes the element on top of the stack and returns that element. */
public int pop() {
if (queue1.isEmpty()) {
while (queue2.size() > 1) {
queue1.offer(queue2.poll());
}
return queue2.poll();
} else {
while (queue1.size() > 1) {
queue2.offer(queue1.poll());
}
return queue1.poll();
}
}
/** * Get the top element. */
public int top() {
int top = 0;
if (queue1.isEmpty()) {
while (queue2.size() > 1) {
queue1.offer(queue2.poll());
}
top = queue2.poll();
queue1.offer(top);
} else {
while (queue1.size() > 1) {
queue2.offer(queue1.poll());
}
top = queue1.poll();
queue2.offer(top);
}
return top;
}
/** * Returns whether the stack is empty. */
public boolean empty() {
return queue2.isEmpty() && queue1.isEmpty();
}
}
柱形图中的最大面积
解法一：暴力枚举高度
外层循环枚举每个节点
内层循环枚举从当前节点开始的后面每个节点，每当枚举一个节点就计算一次最低高度和最大面积
class Solution {
public int largestRectangleArea(int[] heights) {
int maxarea = 0;
for (int i = 0; i < heights.length; i++) {
int minheight = Integer.MAX_VALUE;
for (int j = i; j < heights.length; j++) {
minheight = Math.min(minheight,heights[j]);
maxarea = Math.max(maxarea,minheight*(j-i+1));
}
}
return maxarea;
}
}
解法二：单调栈
官方题解，玩法太高端了，常看看
public class Solution {
public int largestRectangleArea(int[] heights) {
int len = heights.length;
if (len == 0) return 0;
if (len == 1) return heights[0];
int maxSize = 0;
Stack<Integer> stack = new Stack<>();
stack.push(-1);
for (int i = 0; i < len; i++) {
while (stack.peek() != -1 && heights[stack.peek()] >= heights[i]) {
maxSize = Math.max(maxSize, heights[stack.pop()] * (i - stack.peek() - 1));
}
stack.push(i);
}
while (stack.peek() != -1) {
maxSize = Math.max(maxSize, heights[stack.pop()] * (heights.length - stack.peek() - 1));
}
return maxSize;
}
}
接雨水
压栈出栈
总体的原则就是：
-
当前高度小于等于栈顶高度，入栈，指针后移。
-
当前高度大于栈顶高度，出栈，计算出当前墙和栈顶的墙之间水的多少，然后计算当前的高度和新栈的高度的关系，重复第 2 步。直到当前墙的高度不大于栈顶高度或者栈空，然后把当前墙入栈，指针后移。
public class Solution {
public int trap(int[] height) {
int sum = 0;
Stack<Integer> stack = new Stack<>();
int current = 0;
while (current < height.length) {
//如果栈不为空且当前指向的高度大于栈顶高度就一直循环
while (!stack.empty() && height[current] > height[stack.peek()]) {
int h = height[stack.pop()];//要出栈的元素
if (stack.isEmpty()) {//为空就存不了水
break;
}
int distance = current - stack.peek() - 1;//两堵墙之间的距离
int min = Math.min(height[stack.peek()], height[current]);
sum += distance * (min - h);
}
stack.push(current);//当前墙入栈
current++;//指针后移
}
return sum;
}
}
每日温度
运用栈，如果小于栈顶值存入栈，如果大于栈顶的值就将栈顶元素出栈，然后计算距离存入结果数组，反复出栈直到不满足大于栈顶元素
public class Solution {
public int[] dailyTemperatures(int[] T) {
Stack<Integer> stack = new Stack<>();
int[] res = new int[T.length];
for (int i = 0; i < T.length; i++) {
while (!stack.isEmpty() && T[i] > T[stack.peek()]) {
Integer pre = stack.pop();
res[pre] = i-pre;
}
stack.push(i);
}
return res;
}
}
下一个更大元素II
和每日温度类似，只是循环链表用两倍数组长度求模来模拟了
public class Solution {
public int[] nextGreaterElements(int[] nums) {
int len = nums.length;
int[] res = new int[len];
Arrays.fill(res, -1);
Stack<Integer> stack = new Stack<>();
for (int i = 0; i < 2 * len; i++) {
while (!stack.isEmpty() && nums[i % len] > nums[stack.peek() % len]) {
Integer pre = stack.pop();
res[pre % len] = nums[i % len];
}
stack.push(i);
}
return res;
}
}
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