java.util.Stack.add()方法的使用及代码示例

x33g5p2x  于2022-01-29 转载在 其他  
字(10.0k)|赞(0)|评价(0)|浏览(194)

本文整理了Java中java.util.Stack.add()方法的一些代码示例，展示了Stack.add()的具体用法。这些代码示例主要来源于Github/Stackoverflow/Maven等平台，是从一些精选项目中提取出来的代码，具有较强的参考意义，能在一定程度帮忙到你。Stack.add()方法的具体详情如下：
包路径：java.util.Stack
类名称：Stack
方法名：add

Stack.add介绍

暂无

代码示例

代码示例来源：origin: jenkinsci/jenkins

  1. private Set<AbstractProject> getTransitive(Map<AbstractProject, List<DependencyGroup>> direction, AbstractProject src, boolean up) {
  2.   Set<AbstractProject> visited = new HashSet<AbstractProject>();
  3.   Stack<AbstractProject> queue = new Stack<AbstractProject>();
  4.   queue.add(src);
  5.   while(!queue.isEmpty()) {
  6.     AbstractProject p = queue.pop();
  7.     for (AbstractProject child : get(direction,p,up)) {
  8.       if(visited.add(child))
  9.         queue.add(child);
  10.     }
  11.   }
  12.   return visited;
  13. }

代码示例来源：origin: gocd/gocd

  1. private Set<Class> findAllInterfacesInHierarchy(Class candidateGoExtensionClass) {
  2.   Stack<Class> classesInHierarchy = new Stack<>();
  3.   classesInHierarchy.add(candidateGoExtensionClass);
  4.   Set<Class> interfaces = new HashSet<>();
  5.   while (!classesInHierarchy.empty()) {
  6.     Class classToCheckFor = classesInHierarchy.pop();
  7.     if (classToCheckFor.isInterface()) {
  8.       interfaces.add(classToCheckFor);
  9.     }
  10.     classesInHierarchy.addAll(Arrays.asList(classToCheckFor.getInterfaces()));
  11.     if (classToCheckFor.getSuperclass() != null) {
  12.       classesInHierarchy.add(classToCheckFor.getSuperclass());
  13.     }
  14.   }
  15.   return interfaces;
  16. }

代码示例来源：origin: scouter-project/scouter

  1. private static long getEndTime() {
  2.   if(stack.size()==0){
  3.     for(int i = 0 ; i <24 ; i++){
  4.       for(int j=0;j<rate[i] ;j++){
  5.         stack.add(i);
  6.       }
  7.     }
  8.   }
  9.   int h = stack.pop();
  10.   return time + h*3600*1000L +r.nextInt(3600)*1000L;
  11. }

代码示例来源：origin: apache/flink

  1. extractionStates.add(new SharedBufferAccessor.ExtractionState(Tuple2.of(nodeId, entry), version, new Stack<>()));
  2.   final SharedBufferAccessor.ExtractionState extractionState = extractionStates.pop();
  3.       final NodeId currentPathEntry = currentPath.pop().f0;

代码示例来源：origin: scouter-project/scouter

  1. private static long getEndTime() {
  2.   if(stack.size()==0){
  3.     for(int i = 0 ; i <24 ; i++){
  4.       for(int j=0;j<rate[i] ;j++){
  5.         stack.add(i);
  6.       }
  7.     }
  8.   }
  9.   int h = stack.pop();
  10.   return time + h*3600*1000L +r.nextInt(3600)*1000L;
  11. }

代码示例来源：origin: aws/aws-sdk-java

  1. commonPrefixes.add(keyPrefix);
  2. long totalSize = 0;
  3.   String prefix = commonPrefixes.pop();
  4.   ObjectListing listObjectsResponse = null;

代码示例来源：origin: cucumber/cucumber-jvm

  1. private static boolean hasAnnotation(Annotation annotation, Collection<Class<? extends Annotation>> desired) {
  2.   Set<Class<? extends Annotation>> seen = new HashSet<Class<? extends Annotation>>();
  3.   Stack<Class<? extends Annotation>> toCheck = new Stack<Class<? extends Annotation>>();
  4.   toCheck.add(annotation.annotationType());
  5.   while (!toCheck.isEmpty()) {
  6.     Class<? extends Annotation> annotationType = toCheck.pop();
  7.     if (desired.contains(annotationType)) {
  8.       return true;
  9.     }
  10.     seen.add(annotationType);
  11.     for (Annotation annotationTypesAnnotations : annotationType.getAnnotations()) {
  12.       if (!seen.contains(annotationTypesAnnotations.annotationType())) {
  13.         toCheck.add(annotationTypesAnnotations.annotationType());
  14.       }
  15.     }
  16.   }
  17.   return false;
  18. }

代码示例来源：origin: bobbylight/RSyntaxTextArea

  1. /**
  2.  * Returns the first descendant of a component that is an
  3.  * <code>RTextArea</code>.  This is primarily here to support
  4.  * <code>javax.swing.JLayer</code>s that wrap <code>RTextArea</code>s.
  5.  *
  6.  * @param comp The component to recursively look through.
  7.  * @return The first descendant text area, or <code>null</code> if none
  8.  *         is found.
  9.  */
  10. private static RTextArea getFirstRTextAreaDescendant(Component comp) {
  11.   Stack<Component> stack = new Stack<Component>();
  12.   stack.add(comp);
  13.   while (!stack.isEmpty()) {
  14.     Component current = stack.pop();
  15.     if (current instanceof RTextArea) {
  16.       return (RTextArea)current;
  17.     }
  18.     if (current instanceof Container) {
  19.       Container container = (Container)current;
  20.       stack.addAll(Arrays.asList(container.getComponents()));
  21.     }
  22.   }
  23.   return null;
  24. }

代码示例来源：origin: hankcs/HanLP

  1. s.add(new Integer[]{1, -1});
  2.   if (childList == null || (childList.size() - 1) == pair[1])
  3.     s.pop();
  4.     if (s.empty())
  5.   if (base[c] > 0)
  6.     s.add(new Integer[]{base[c], -1});
  7.     charBuf.add(code);
  8.     continue;

代码示例来源：origin: aws-amplify/aws-sdk-android

  1. commonPrefixes.add(keyPrefix);
  2. long totalSize = 0;
  3.   final String prefix = commonPrefixes.pop();
  4.   ObjectListing listObjectsResponse = null;

代码示例来源：origin: stanfordnlp/CoreNLP

  1. public TreeNode<E,T> balance(TreeNode<E,T> node) {
  2.  if (debug) check(node);
  3.  Stack<TreeNode<E,T>> todo = new Stack<>();
  4.  todo.add(node);
  5.  TreeNode<E,T> newRoot = null;
  6.  while (!todo.isEmpty()) {
  7.   TreeNode<E,T> n = todo.pop();
  8.   // Balance tree between this node
  9.   // Select median nodes and try to balance the tree
  10.   int medianAt = n.size/2;
  11.   TreeNode<E,T> median = getNode(n, medianAt);
  12.   // Okay, this is going to be our root
  13.   if (median != null && median != n) {
  14.    // Yes, there is indeed something to be done
  15.    rotateUp(median, n);
  16.   }
  17.   if (newRoot == null) {
  18.    newRoot = median;
  19.   }
  20.   if (median.left != null) todo.push(median.left);
  21.   if (median.right != null) todo.push(median.right);
  22.  }
  23.  if (newRoot == null) return node;
  24.  else return newRoot;
  25. }

代码示例来源：origin: facebook/litho

  1. @Override
  2. public List<InspectableComponent> extract(LithoView lithoView) {
  3.  final List<InspectableComponent> res = new LinkedList<>();
  4.  final Stack<InspectableComponent> stack = new Stack<>();
  5.  final InspectableComponent rootInstance = InspectableComponent.getRootInstance(lithoView);
  6.  Preconditions.checkNotNull(
  7.    rootInstance,
  8.    "Could not obtain DebugComponent. "
  9.      + "Please ensure that ComponentsConfiguration.IS_INTERNAL_BUILD is enabled.");
  10.  stack.add(rootInstance);
  11.  while (!stack.isEmpty()) {
  12.   final InspectableComponent inspectableComponent = stack.pop();
  13.   res.add(inspectableComponent);
  14.   for (InspectableComponent child : inspectableComponent.getChildComponents()) {
  15.    stack.push(child);
  16.   }
  17.  }
  18.  return res;
  19. }

代码示例来源：origin: stanfordnlp/CoreNLP

  1. Integer index = fileArrayStackIndices.pop();
  2.  int ind = index.intValue();
  3.  if (ind < files.length) {
  4.   fileArrayStack.pop();
  5.  fileArrayStack.pop();
  6.  if (obj instanceof String) {
  7.   obj = new File((String) obj);
  8. fileArrayStack.add(roots[rootsIndex]);
  9. fileArrayStackIndices.push(Integer.valueOf(0));

代码示例来源：origin: stanfordnlp/CoreNLP

  1. public static <E extends Comparable<E>, T extends HasInterval<E>> boolean overlaps(TreeNode<E,T> node, Interval<E> target) {
  2.  Stack<TreeNode<E,T>> todo = new Stack<>();
  3.  todo.push(node);
  4.  while (!todo.isEmpty()) {
  5.   TreeNode<E,T> n = todo.pop();
  6.   // Don't search nodes that don't exist
  7.   if (n == null || n.isEmpty()) continue;
  8.   // If target is to the right of the rightmost point of any interval
  9.   // in this node and all children, there won't be any matches.
  10.   if (target.first.compareTo(n.maxEnd) > 0)
  11.     continue;
  12.   // Check this node
  13.   if (n.value.getInterval().overlaps(target)) {
  14.     return true;
  15.   }
  16.   // Search left children
  17.   if (n.left != null) {
  18.    todo.add(n.left);
  19.   }
  20.   // If target is to the left of the start of this interval,
  21.   // then it can't be in any child to the right.
  22.   if (target.second.compareTo(n.value.getInterval().first()) < 0)  {
  23.    continue;
  24.   }
  25.   if (n.right != null)  {
  26.    todo.add(n.right);
  27.   }
  28.  }
  29.  return false;
  30. }

代码示例来源：origin: k9mail/k-9

  1. static Part findPartById(Part searchRoot, long partId) {
  2.   if (searchRoot instanceof LocalMessage) {
  3.     LocalMessage localMessage = (LocalMessage) searchRoot;
  4.     if (localMessage.getMessagePartId() == partId) {
  5.       return localMessage;
  6.     }
  7.   }
  8.   Stack<Part> partStack = new Stack<>();
  9.   partStack.add(searchRoot);
  10.   while (!partStack.empty()) {
  11.     Part part = partStack.pop();
  12.     if (part instanceof LocalPart) {
  13.       LocalPart localBodyPart = (LocalPart) part;
  14.       if (localBodyPart.getPartId() == partId) {
  15.         return part;
  16.       }
  17.     }
  18.     Body body = part.getBody();
  19.     if (body instanceof Multipart) {
  20.       Multipart innerMultipart = (Multipart) body;
  21.       for (BodyPart innerPart : innerMultipart.getBodyParts()) {
  22.         partStack.add(innerPart);
  23.       }
  24.     }
  25.     if (body instanceof Part) {
  26.       partStack.add((Part) body);
  27.     }
  28.   }
  29.   return null;
  30. }

代码示例来源：origin: pxb1988/dex2jar

  1. stack.add(method.stmts.getFirst());
  2. while (!stack.isEmpty()) {
  3.   Stmt currentStmt = stack.pop();
  4.   if (currentStmt.visited) {
  5.     continue;
  6.     Collections.addAll(stack, bs.targets);
  7.     LabelStmt target = bs.defaultTarget;
  8.     stack.add(target);
  9.     stack.add(target);
  10.     stack.add(target);

代码示例来源：origin: pedrovgs/Algorithms

  1. /**
  2.  * Iterative implementation of this binary tree traversal. The complexity order in time terms of
  3.  * this algorithm is O(N) where N is the number of nodes in the tree. In space terms the
  4.  * complexity order of this algorithm is also O(N) where N is the number of nodes we have to
  5.  * store in the auxiliary data structure, the stack.
  6.  */
  7. public List<BinaryNode> getIterative(BinaryNode root) {
  8.  validateBinaryNode(root);
  9.  List<BinaryNode> result = new LinkedList<BinaryNode>();
  10.  Stack<BinaryNode> stack = new Stack<BinaryNode>();
  11.  stack.push(root);
  12.  while (!stack.isEmpty()) {
  13.   BinaryNode node = stack.pop();
  14.   result.add(node);
  15.   if (node.hasRight()) {
  16.    stack.add(node.getRight());
  17.   }
  18.   if (node.hasLeft()) {
  19.    stack.add(node.getLeft());
  20.   }
  21.  }
  22.  return result;
  23. }

代码示例来源：origin: Alluxio/alluxio

  1. /**
  2.  * @param source an Alluxio URI
  3.  * @param fileSystem the Alluxio file system
  4.  * @return whether the URI is a file or a directory which contains files (including recursively)
  5.  * @throws Exception if an unexpected exception occurs
  6.  */
  7. private static boolean hasFiles(AlluxioURI source, FileSystem fileSystem) throws Exception {
  8.  Stack<AlluxioURI> dirsToCheck = new Stack<>();
  9.  dirsToCheck.add(source);
  10.  while (!dirsToCheck.isEmpty()) {
  11.   try {
  12.    for (URIStatus status : fileSystem.listStatus(dirsToCheck.pop())) {
  13.     if (!status.isFolder()) {
  14.      return true;
  15.     }
  16.     dirsToCheck.push(new AlluxioURI(status.getPath()));
  17.    }
  18.   } catch (FileDoesNotExistException e) {
  19.    // This probably means another worker has deleted the directory already, so we can probably
  20.    // return false here. To be safe though, we will fall through and complete the search.
  21.   }
  22.  }
  23.  return false;
  24. }

代码示例来源：origin: apache/incubator-pinot

  1. /**
  2.  * Extracts all columns from the given filter query tree.
  3.  */
  4. public static Set<String> extractFilterColumns(FilterQueryTree root) {
  5.  Set<String> filterColumns = new HashSet<>();
  6.  if (root.getChildren() == null) {
  7.   filterColumns.add(root.getColumn());
  8.  } else {
  9.   Stack<FilterQueryTree> stack = new Stack<>();
  10.   stack.add(root);
  11.   while (!stack.empty()) {
  12.    FilterQueryTree node = stack.pop();
  13.    for (FilterQueryTree child : node.getChildren()) {
  14.     if (child.getChildren() == null) {
  15.      filterColumns.add(child.getColumn());
  16.     } else {
  17.      stack.push(child);
  18.     }
  19.    }
  20.   }
  21.  }
  22.  return filterColumns;
  23. }

代码示例来源：origin: stanfordnlp/CoreNLP

  1. public void check(TreeNode<E,T> treeNode) {
  2.  Stack<TreeNode<E,T>> todo = new Stack<>();
  3.  todo.add(treeNode);
  4.  while (!todo.isEmpty()) {
  5.   TreeNode<E,T> node = todo.pop();
  6.   if (node == node.parent) {
  7.    throw new IllegalStateException("node is same as parent!!!");
  8.   if (node.left != null) todo.add(node.left);
  9.   if (node.right != null) todo.add(node.right);

内容来源于网络，如有侵权，请联系作者删除！

stack

相关文章

    查看更多

    热门标签

    更多
    JavaquerypythonNode开发语言requestUtil数据库Table后端算法LoggerMessageElementParser

    最新文章

    更多
    • 蜀ICP备13028337号-1 大数据知识库 https://www.saoniuhuo.com © All rights reserved
    • 本站内容来源互联网,如果侵犯您的权益请联系我们删除, 联系方式：448109455@qq.com