Leetcode——队列的最大值

Posted 2021-08-20 Yawn,

1. 题目

2. 题解

（1）暴力

直接实现一个普通的队列，查询最大值时遍历计算

class MaxQueue {
    int[] q = new int[20000];
    int begin = 0, end = 0;

    public MaxQueue() {

    }
    
    public int max_value() {
        int ans = -1;
        for (int i = begin; i != end; ++i) {
            ans = Math.max(ans, q[i]);
        }
        return ans;
    }
    
    public void push_back(int value) {
        q[end++] = value;
    }
    
    public int pop_front() {
        if (begin == end) {
            return -1;
        }
        return q[begin++];
    }
}

（2）单调队列

• 两个双端队列，一个存储队列，一个存储最大值
• 始终保持最大值队列为降序队列
• 注意：出队始终是从前往后
• 如下图给出的解释
  

class MaxQueue {
    Deque<Integer> deque;
    Deque<Integer> maxDeque;
    public MaxQueue() {
        deque = new LinkedList<>();
        maxDeque = new LinkedList<>();
        //maxDeque.addLast(Integer.MIN_VALUE);
    }
    
    public int max_value() {
        if(maxDeque.isEmpty())
            return -1;
        return maxDeque.peekFirst();
    }
    
    public void push_back(int value) {
        //保持单调队列为降序
        while(!maxDeque.isEmpty() && maxDeque.peekLast() < value)
            maxDeque.pollLast();
        maxDeque.addLast(value);
        deque.add(value);
    }
    
    public int pop_front() {
        if(deque.isEmpty())
            return -1;
        int val = deque.poll();
        if(val == maxDeque.peekFirst())
            maxDeque.pollFirst();
        return val;
    }
}

/**
 * Your MaxQueue object will be instantiated and called as such:
 * MaxQueue obj = new MaxQueue();
 * int param_1 = obj.max_value();
 * obj.push_back(value);
 * int param_3 = obj.pop_front();
 */