代码随想录四刷day11

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前言

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class MyQueue { Deque<Integer> deque = new LinkedList<>(); //弹出元素时，比较当前要弹出的数值是否等于队列出口的数值，如果相等则弹出 //同时判断队列当前是否为空 void poll(int val) { if (!deque.isEmpty() && val == deque.peek()) { deque.poll(); } } //添加元素时，如果要添加的元素大于入口处的元素，就将入口元素弹出 //保证队列元素单调递减 //比如此时队列元素3,1，2将要入队，比1大，所以1弹出，此时队列：3,2 void add(int val) { while (!deque.isEmpty() && val > deque.getLast()) { deque.removeLast(); } deque.add(val); } //队列队顶元素始终为最大值 int peek() { return deque.peek(); } }

一、力扣20. 有效的括号

class Solution {
    Deque<Character> deq = new LinkedList<>();
    public boolean isValid(String s) {
        for(char c : s.toCharArray()){
            if(c == '(' || c == '{' || c == '['){
                deq.offerLast(c);
            }else{
                if(deq.isEmpty()){
                    return false;
                }
                char temp = deq.pollLast();
                if(c == ')' && temp != '('){
                    return false;
                }
                if(c == '}' && temp != '{'){
                    return false;
                }
                if(c == ']' && temp != '['){
                    return false;
                }
            }
        }
        if(!deq.isEmpty()){
            return false;
        }
        return true;
    }
}

二、力扣1047. 删除字符串中的所有相邻重复项

class Solution {
    public String removeDuplicates(String s) {
        StringBuilder sb = new StringBuilder();
        for(char c : s.toCharArray()){
            if(sb.length() == 0){
                sb.append(c);
            }else{
                if(sb.charAt(sb.length()-1) == c){
                    sb.deleteCharAt(sb.length()-1);
                }else{
                    sb.append(c);
                }
            }
        }
        return sb.toString();
    }
}

三、力扣150. 逆波兰表达式求值

class Solution {
    public int evalRPN(String[] tokens) {
        Deque<Integer> deq = new LinkedList<>();
        for(String s : tokens){
            if(s.equals("+") || s.equals("-") || s.equals("*") || s.equals("/")){
                int right = deq.pollLast();
                int left = deq.pollLast();
                switch(s){
                    case "+":
                        deq.offerLast(left+right);break;
                    case "-":
                        deq.offerLast(left-right);break;
                    case "*":
                        deq.offerLast(left*right);break;
                    case "/":
                        deq.offerLast(left/right);break;
                }
            }else{
                deq.offerLast(Integer.parseInt(s));
            }
        }
        return deq.pollLast();
    }
}

四、力扣239. 滑动窗口最大值

class Solution {
    Deque<Integer> deq = new LinkedList<>();
    public int[] maxSlidingWindow(int[] nums, int k) {
        int[] res = new int[nums.length - k +1];
        int left = 0, right = 0, t = 0;
        while(right < nums.length){
            if(right < k){
                if(!deq.isEmpty()){
                    if(deq.peekLast() == nums[right]){
                        deq.offerLast(nums[right]);
                    }else if(deq.peekLast() < nums[right]){
                        while(!deq.isEmpty() && nums[right] > deq.peekLast()){
                            deq.pollLast();
                        }
                        deq.offerLast(nums[right]);
                    }else{
                        deq.offerLast(nums[right]);
                    }
                }else{
                    deq.offerLast(nums[right]);
                }
                if(right == k-1){
                    res[t ++] = deq.peekFirst();
                    if(deq.peekFirst() == nums[left]){
                        deq.pollFirst();
                    }
                    left ++;
                }
                right ++;
                
            }else{
                
                if(!deq.isEmpty()){
                    if(deq.peekLast() == nums[right]){
                        deq.offerLast(nums[right]);
                    }else if(deq.peekLast() < nums[right]){
                        while(!deq.isEmpty()  && nums[right] > deq.peekLast()){
                            deq.pollLast();
                        }
                        deq.offerLast(nums[right]);
                    }else{
                        deq.offerLast(nums[right]);
                    }
                }else{
                    deq.offerLast(nums[right]);
                }
                res[t ++] = deq.peekFirst();
                right ++;
                if(deq.peekFirst() == nums[left]){
                    deq.pollFirst();
                }
                left ++;
            }
        }
        return res;
    }
}