packageDate_pacage;publicclassLinkedListStack<E>implementsStack<E>{publicstaticvoidmain(String[] args) { LinkedListStack<Integer> stack =newLinkedListStack<>();for(inti = 0 ; i < 5 ; i ++) { stack.push(i); System.out.println(stack); } stack.pop(); System.out.println(stack); }privat...
首先,我们需要创建一个LinkedList对象来表示栈: LinkedList<Integer>stack=newLinkedList<>(); 1. 接下来,我们可以使用push()方法向栈中添加元素: stack.push(1);stack.push(2);stack.push(3); 1. 2. 3. 最后,我们可以使用pop()方法从栈中移除并返回栈顶元素: inttop=stack.pop(); 1. 4. 创建队列 ...
// 使用 LinkedList 作为栈 LinkedList<Integer> stack = new LinkedList<>(); stack.push(10); // 添加到栈顶 stack.push(20); System.out.println(stack.pop()); // 输出 20 // 使用 LinkedList 作为队列 LinkedList<Integer> queue = new LinkedList<>(); queue.offer(10); // 添加到队列尾部 que...
Deque<Integer> stack = new LinkedList<>(); stack.push(1); stack.push(2); stack.push(3); while (stack.peek() != null) { System.out.println(stack.pop()); } /** * output: * 3 * 2 * 1 */ 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. Java中还有一个Stack类,...
LinkedList<String> stack=newLinkedList<>();//push(E e) 将元素推入此列表所表示的堆栈。是靠着addFirst(e)实现;stack.push("1"); stack.push("2"); stack.push("3"); stack.push("4"); log.info("test LinkedList"); log.info(stack.toString());//[4, 3, 2, 1]//peek() 获取但不移除...
lst = new LinkedList<LogEvent>(); final Map<Integer, List<LogEvent>> map = new HashMap...
stack = (LinkedList) Collections.synchronizedList(stack); 10. } 11. } 17、实现队列 [ java] view plaincopy 1. import java.util.LinkedList; 2. public class Main { 3. public static void main(String[] argv) throws Exception { 4. LinkedList queue = new LinkedList(); 5. Object object = "...
问LinkedList<E>中pop()、remove()和poll()之间的真正区别EN(1)select==>时间复杂度O(n) 仅知道...
NewLinkedList() values := []int{5, 7, 12, 9} for _, v := range values { ll.Add(v) } ll.AddTo(2, 18) v3, _ := ll.Remove(3) fmt.Printf("ll.Remove(3) = %v\n", v3) // Iterate all the elements fmt.Println("Iterate: ") it, hasNext := ll.Iterator() var v ...
Stack Heap head 1 2 tail newNode 3 This is just a special case of the general rule: to insert or delete a node inside a list, you need a pointer to the node just before that position, so you can change its .next field. Many list problems include the sub-problem of advancing a ...