1. 滑动窗口的最大值

给定一个数组 nums 和滑动窗口的大小 k，请找出所有滑动窗口里的最大值。

示例:
输入: nums = [1,3,-1,-3,5,3,6,7], 和 k = 3
输出: [3,3,5,5,6,7]
解释:

  滑动窗口的位置                最大值
---------------               -----
[1  3  -1] -3  5  3  6  7       3
 1 [3  -1  -3] 5  3  6  7       3
 1  3 [-1  -3  5] 3  6  7       5
 1  3  -1 [-3  5  3] 6  7       5
 1  3  -1  -3 [5  3  6] 7       6
 1  3  -1  -3  5 [3  6  7]      7

1. 大根堆

    public ArrayList<Integer> maxInWindows(int [] num, int size) {
        ArrayList<Integer> list = new ArrayList<>();
        if (num == null || num.length == 0 || size < 1) return list;
        PriorityQueue<Integer> queue = new PriorityQueue<>((t1, t2) -> t2 - t1);
        for (int i = 0; i < num.length; i++) {
            // 左边界
            int j = i + 1 - size;
            queue.offer(num[i]);
            // 左边滑出
            if (j > 0 && num[j - 1] == queue.peek()) {
                queue.poll();
            }
            // 左边界
            if (j >= 0) list.add(queue.peek());
        }
        return list;
    }

2. 记录最大项的位置

public class Solution {
    public ArrayList<Integer> maxInWindows(int [] num, int size) {
        ArrayList<Integer> list = new ArrayList<>();
        if (num == null || num.length == 0 || size < 1) return list;
        int maxIndex = -1, max = Integer.MIN_VALUE;;
        for (int i = 0; i < num.length; i++) {
            // 左边界
            int j = i + 1 - size;
            if (num[i] >= max) {
                max = num[i];
                maxIndex = i;
            }
            // 左边滑出
            if (maxIndex < j) {
                maxIndex = findMax(num, j, i);
                max = num[maxIndex];
            }
            // 左边界
            if (j >= 0) list.add(max);
        }
        return list;
    }
    
    private int findMax(int[] num, int i, int j) {
        int index = i;
        int max = num[i];
        for (int k = i; k < j; k++) {
            if (num[k] >= max) {
                index = k;
                max = num[k];
            }
        }
        return index;
    }
}

2.队列的最大值

请定义一个队列并实现函数 max_value 得到队列里的最大值，要求函数max_value、push_back 和 pop_front 的均摊时间复杂度都是O(1)。

若队列为空，pop_front 和 max_value 需要返回 -1
输入:

["MaxQueue","push_back","push_back","max_value","pop_front","max_value"]
[[],[1],[2],[],[],[]]
输出: [null,null,null,2,1,2]

class MaxQueue {

    private int maxValue = Integer.MIN_VALUE;
    private LinkedList<Integer> queue;

    public MaxQueue() {
        queue = new LinkedList<>();
    }
    
    public int max_value() {
        if (queue.size() == 0) return -1;
        return maxValue;
    }
    
    public void push_back(int value) {
        queue.addLast(value);
        if (value > maxValue) {
            maxValue = value;
        }
    }
    
    public int pop_front() {
        if (queue.size() == 0) return -1;
        int first = queue.removeFirst();
        if (queue.size() == 0) {
            maxValue = Integer.MIN_VALUE;
            return first;
        }
        if (first == maxValue) {
            maxValue = Integer.MIN_VALUE;
            Iterator<Integer> i = queue.iterator();
            while(i.hasNext()) {
                int tmp = i.next();
                if (maxValue < tmp) {
                    maxValue = tmp;
                }
            }
        }
        return first;
    }
}

/**
 * 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();
 */