zoukankan      html  css  js  c++  java
  • 遍历Map的四种方法

    public static void main(String[] args) {


      Map<String, String> map = new HashMap<String, String>();
      map.put("1", "value1");
      map.put("2", "value2");
      map.put("3", "value3");
      
      //第一种:普遍使用,二次取值
      System.out.println("通过Map.keySet遍历key和value:");
      for (String key : map.keySet()) {
       System.out.println("key= "+ key + " and value= " + map.get(key));
      }
      
      //第二种
      System.out.println("通过Map.entrySet使用iterator遍历key和value:");
      Iterator<Map.Entry<String, String>> it = map.entrySet().iterator();
      while (it.hasNext()) {
       Map.Entry<String, String> entry = it.next();
       System.out.println("key= " + entry.getKey() + " and value= " + entry.getValue());
      }
      
      //第三种:推荐,尤其是容量大时
      System.out.println("通过Map.entrySet遍历key和value");
      for (Map.Entry<String, String> entry : map.entrySet()) {
       System.out.println("key= " + entry.getKey() + " and value= " + entry.getValue());
      }

      //第四种
      System.out.println("通过Map.values()遍历所有的value,但不能遍历key");
      for (String v : map.values()) {
       System.out.println("value= " + v);
      }
     }

  • 相关阅读:
    771. Jewels and Stones
    706. Design HashMap
    811. Subdomain Visit Count
    733. Flood Fill
    117. Populating Next Right Pointers in Each Node II
    250. Count Univalue Subtrees
    94. Binary Tree Inorder Traversal
    116. Populating Next Right Pointers in Each Node
    285. Inorder Successor in BST
    292. Nim Game Java Solutin
  • 原文地址:https://www.cnblogs.com/yhtboke/p/5690459.html
Copyright © 2011-2022 走看看