备战秋招day6

算法

232. 用栈实现队列

class MyQueue {Stack<Integer> in;Stack<Integer> out;//构造器public MyQueue() {in = new Stack<>();out = new Stack<>();}//入队到末尾public void push(int x) {in.push(x);}//出队public int pop() {//先进先出//in的元素放入out里面然后pop就能达到先进先出的目的，元素顺序都被调转了change();return out.pop();}//返回队列开头元素public int peek() {change();return out.peek();}//判空public boolean empty() {return in.isEmpty() && out.isEmpty();}//写一个方法：做将元素从in转到outpublic void change(){//出队的首要条件是：out为空if(!out.isEmpty()) return;while(!in.isEmpty()){out.push(in.pop());}}
}

225. 用队列实现栈

class MyStack {Deque<Integer> que;public MyStack() {que = new ArrayDeque<>();}//每次都加到最前面（只能使用标准操作）public void push(int x) {que.push(x);int size = que.size();//我们无需移动刚进来的数while(size-- >1){que.push(que.pop());}}//先进后出:由于我们的添加操作，导致最先进来的一定在最左边public int pop() {return que.pop();}public int top() {return que.peek();}public boolean empty() {return que.isEmpty();}
}

20. 有效的括号

class Solution {public boolean isValid(String s) {//用栈来做：思路就跟消消乐一样Stack<Character> stack = new Stack<>();int len = s.length();for(int i = 0;i<len;i++){char ch = s.charAt(i);if(ch == '('){stack.push(')');}else if(ch == '['){stack.push(']');}else if(ch == '{'){stack.push('}');}else if(stack.isEmpty() || stack.peek()!=ch){return false;}else{stack.pop();}}return stack.isEmpty();}
}

1047. 删除字符串中的所有相邻重复项

class Solution {public String removeDuplicates(String s) {//依旧用栈来做Stack<Character> stack = new Stack<>();int len = s.length();for(int i = 0;i<len;i++){char ch = s.charAt(i);//相同则出栈if(!stack.isEmpty() && stack.peek()==ch){char ca = stack.pop();//System.out.println("pop: " + ca + ": " + i);continue;}stack.push(ch);//System.out.println("push: " + ch + ": " + i);}String res = "";while(!stack.isEmpty()){res = stack.pop() + res;//System.out.println(res);}return res;}
}

150. 逆波兰表达式求值

class Solution {public int evalRPN(String[] tokens) {//很显然使用栈来解决Stack<Integer> stack = new Stack<>();for(String s : tokens){if(s.equals("+")){stack.push(stack.pop() + stack.pop());}else if(s.equals("-")){stack.push(-stack.pop() + stack.pop());}else if(s.equals("*")){stack.push(stack.pop() * stack.pop());}else if(s.equals("/")){int temp1 = stack.pop();int temp2 = stack.pop();stack.push(temp2 / temp1);}else{//转换值stack.push(Integer.valueOf(s));}}return stack.pop();}
}

239. 滑动窗口最大值

class Solution {//为什么说这道题是单调队列？//很明显，在几个窗口里面，我们需要维护这些窗口里面的一个单调值（这道题来说就是最大值）//利用单调队列来做，可以实现入队出队，以及维护我们需要的这个单调情况这些操作//而java里面可以通过双端队列来实现public int[] maxSlidingWindow(int[] nums, int k) {Deque<Integer> deque = new LinkedList<>();int n = nums.length;//有n个数，只能看到k个数字，res只需要最多n-k个即可int[] res = new int[n-k+1];int index = 0;//由于我们不好直接把值放入，因此选择把下标放入队列中for(int i = 0;i<n;i++){//不在窗口内的元素就要出队了while(!deque.isEmpty() && deque.peek() < i-k+1){deque.poll();}//与队列内的元素进行比较，保证单调递增while(!deque.isEmpty() && nums[deque.peekLast()] < nums[i]){deque.pollLast();}deque.offer(i);//开始记录最值if(i>=k-1){res[index++] = nums[deque.peek()];}}return res;}
}

347. 前 K 个高频元素

class Solution {public int[] topKFrequent(int[] nums, int k) {//利用Map存元素和次数，然后利用大顶堆或小顶堆来实现Map<Integer,Integer> map = new HashMap<>();for(int i : nums){map.put(i,map.getOrDefault(i,0)+1);}//大顶堆的写法PriorityQueue<int[]> pq = new PriorityQueue<>((pair1, pair2) -> pair2[1] - pair1[1]);for(Map.Entry<Integer,Integer> entry : map.entrySet()){pq.add(new int[]{entry.getKey(),entry.getValue()});}int[]res = new int[k];for(int i = 0;i<k;i++){res[i] = pq.poll()[0];}return res;}
}