Java Code Examples for java.util.ArrayDeque#pollLast()

The following examples show how to use java.util.ArrayDeque#pollLast() . These examples are extracted from open source projects. You can vote up the ones you like or vote down the ones you don't like, and go to the original project or source file by following the links above each example. You may check out the related API usage on the sidebar.
Example 1
/**
 * 滑动窗口应当是队列,但为了得到滑动窗口的最大值,队列序可以从两端删除元素,因此使用双端队列。 对新来的元素k,将其与双端队列中的元素相比较
 * 1)前面比k小的,直接移出队列(因为不再可能成为后面滑动窗口的最大值了!), 2)前面比k大的X,比较两者下标,判断X是否已不在窗口之内,不在了,直接移出队列
 * 队列的第一个元素是滑动窗口中的最大值
 */
public ArrayList<Integer> maxInWindows(int[] num, int size) {
    ArrayList<Integer> list = new ArrayList<>();
    if (size == 0) return list;
    int start = 0;
    // 用来保存可能是滑动窗口最大值的数字的下标
    ArrayDeque<Integer> index = new ArrayDeque<>();
    for (int i = 0; i < num.length; i++) {
        start = i - size + 1;
        if (index.isEmpty()) index.add(i);
        // 如果队列的头部元素已经从滑动窗口里滑出,滑出的数字需要从队列的头部删除
        else if (start > index.peekFirst()) index.pollFirst();
        // 数组:{2,3,4,2,6,2,5,1}
        // 如果已有数字小于待存入的数据, 这些数字已经不可能是滑动窗口的最大值
        // 因此它们将会依次地从队尾删除
        while ((!index.isEmpty()) && num[index.peekLast()] <= num[i]) index.pollLast();
        index.add(i);
        if (start >= 0) list.add(num[index.peekFirst()]);
    }
    return list;
}
 
Example 2
/**
 * Builds a sorted list of views. The sorting order depends on the dependencies
 * between the view. For instance, if view C needs view A to be processed first
 * and view A needs view B to be processed first, the dependency graph
 * is: B -> A -> C. The sorted array will contain views B, A and C in this order.
 *
 * @param sorted The sorted list of views. The length of this array must
 *        be equal to getChildCount().
 * @param rules The list of rules to take into account.
 */
void getSortedViews(View[] sorted, int... rules) {
    final ArrayDeque<Node> roots = findRoots(rules);
    int index = 0;

    Node node;
    while ((node = roots.pollLast()) != null) {
        final View view = node.view;
        final int key = view.getId();

        sorted[index++] = view;

        final ArrayMap<Node, DependencyGraph> dependents = node.dependents;
        final int count = dependents.size();
        for (int i = 0; i < count; i++) {
            final Node dependent = dependents.keyAt(i);
            final SparseArray<Node> dependencies = dependent.dependencies;

            dependencies.remove(key);
            if (dependencies.size() == 0) {
                roots.add(dependent);
            }
        }
    }

    if (index < sorted.length) {
        throw new IllegalStateException("Circular dependencies cannot exist"
                + " in RelativeLayout");
    }
}
 
Example 3
public int[] maxSlidingWindow(int[] nums, int k) {
    int len = nums.length;
    // 特判
    if (len == 0) {
        return new int[]{};
    }
    // 结果集
    List<Integer> res = new ArrayList<>();
    // 滑动窗口,注意:保存的是索引值
    ArrayDeque<Integer> deque = new ArrayDeque<>(k);

    for (int i = 0; i < len; i++) {
        // 当元素从左边界滑出的时候,如果它恰恰好是滑动窗口的最大值
        // 那么将它弹出
        if (i >= k && i - k == deque.getFirst()) {
            deque.pollFirst();
        }

        // 如果滑动窗口非空,新进来的数比队列里已经存在的数还要大
        // 则说明已经存在数一定不会是滑动窗口的最大值(它们毫无出头之日)
        // 将它们弹出
        while (!deque.isEmpty() && nums[deque.peekLast()] <= nums[i]) {
            deque.pollLast();
        }
        deque.add(i);
        // 队首一定是滑动窗口的最大值的索引
        if (i >= k - 1) {
            res.add(nums[deque.peekFirst()]);
        }
    }

    int size = res.size();
    int[] result = new int[size];

    for (int i = 0; i < size; i++) {
        result[i] = res.get(i);
    }
    return result;
}
 
Example 4
/**
 * peekFirst() returns element inserted with push
 */
public void testPush() {
    ArrayDeque q = populatedDeque(3);
    q.pollLast();
    q.push(four);
    assertSame(four, q.peekFirst());
}
 
Example 5
Source Project: j2objc   File: ArrayDequeTest.java    License: Apache License 2.0 5 votes vote down vote up
/**
 * peekFirst() returns element inserted with push
 */
public void testPush() {
    ArrayDeque q = populatedDeque(3);
    q.pollLast();
    q.push(four);
    assertSame(four, q.peekFirst());
}
 
Example 6
Source Project: swift-t   File: TaskQueue.java    License: Apache License 2.0 5 votes vote down vote up
/**
 * TODO: each thread should hold local reference to own deque
 * @return a task, or null if nothing available
 */
public Task getTask(int threadNum) {
  // Targeted have highest priority
  Task res = targeted.get(threadNum).pollLast();
  if (res != null)
    return res;
  
  // Next, try to see if something in local deque
  ArrayDeque<Task> myDeque = regular.get(threadNum);
  synchronized (myDeque) {
    res = myDeque.pollLast();
  }
  if (res != null)
    return res;
  
  // Finally, search other deques to steal work
  // TODO: exit condition - detect idle, or go to sleep
  Random r = new Random();
  while (true) {
    int deque = r.nextInt(numThreads - 1);
    if (deque >= threadNum)
      deque++;
    
    ArrayDeque<Task> otherDeque = regular.get(threadNum);
    // Task from other end
    synchronized (otherDeque) {
      res = otherDeque.pollFirst();
    }
    if (res != null)
      return res;      
  }
}