当前位置: 移动技术网 > IT编程>开发语言>Java > java 遍历Map及Map转化为二维数组的实例

java 遍历Map及Map转化为二维数组的实例

2019年07月19日  | 移动技术网IT编程  | 我要评论

钦州学院怎么样,礼仪电子书,歌手突围赛排名

java 遍历map及map转化为二维数组的实例

实例代码:

import java.util.hashmap; 
import java.util.iterator; 
import java.util.map; 
 
public class test { 
  public static void main(string[] args) { 
    int a = 0, b = 0, c = 0; 
    // 第一种:通过map.keyset()遍历map及将map转化为二维数组 
    map<string, string> map1 = new hashmap<string, string>(); 
    map1.put("012013012013", "张三"); 
    map1.put("012013012014", "张四"); 
    string[][] group1 = new string[map1.size()][2]; 
    system.out.println("第一种:通过map.keyset()遍历map1的key和value"); 
    for (string key : map1.keyset()) {  
      system.out.println("key = " + key + " and value = " + map1.get(key));  
      group1[a][0] = key; 
      group1[a][1] = map1.get(key); 
      a++; 
    }  
    system.out.println("map1.size()为:" + map1.size() + ",a为:" + a + ",group1数组的长度为:" + group1.length); 
    system.out.println("----------------------------------------------------"); 
    for(int n = 0; n < group1.length; n++) { 
      system.out.println("key = " + group1[n][0] + " and value = " + group1[n][1]);  
    } 
     
    // 第二种:通过map.entryset()使用iterator()遍历map及将map转化为二维数组 
    map<string, string> map2 = new hashmap<string, string>(); 
    map2.put("112013012013", "李三"); 
    map2.put("112013012014", "李四"); 
    system.out.println("\n" + "第二种:通过map.entryset()使用iterator()遍历map2的key和value"); 
    iterator<map.entry<string, string>> iterator = map2.entryset().iterator();  
    string[][] group2 = new string[map2.size()][2]; 
    while (iterator.hasnext()) {  
      map.entry<string, string> entry = iterator.next();         
      system.out.println("key = " + entry.getkey() + " and value = " + entry.getvalue()); 
      group2[b][0] = entry.getkey(); 
      group2[b][1] = entry.getvalue(); 
      b++; 
    }  
    system.out.println("map2.size()为:" + map2.size() + ",b为:" + b + ",group2数组的长度为:" + group2.length); 
    system.out.println("----------------------------------------------------"); 
    for(int n = 0; n < group2.length; n++) { 
      system.out.println("key = " + group2[n][0] + " and value = " + group2[n][1]);  
    } 
     
    // 第三种:通过map.entryset()遍历遍历map及将map转化为二维数组 
    map<string, string> map = new hashmap<string, string>(); 
    map.putall(map1); 
    map.putall(map2);   
    string[][] group3 = new string[map.size()][2]; 
    system.out.println("\n" + "第三种:通过map.entryset()遍历map的key和value ");    
    for (map.entry<string, string> entry : map.entryset()) {  
      system.out.println("key = " + entry.getkey() + " and value = " + entry.getvalue());  
      group3[c][0] = entry.getkey(); 
      group3[c][1] = entry.getvalue(); 
      c++; 
    } 
    system.out.println("map.size()为:" + map.size() + ",c为:" + c + ",group3数组的长度为:" + group3.length); 
    system.out.println("----------------------------------------------------"); 
    for(int n = 0; n < group3.length; n++) { 
      system.out.println("key = " + group3[n][0] + " and value = " + group3[n][1]);  
    } 
     
  } 
} 

输出结果为:

第一种:通过map.keyset()遍历map1的key和value 
key = 012013012013 and value = 张三 
key = 012013012014 and value = 张四 
map1.size()为:2,a为:2,group1数组的长度为:2 
---------------------------------------------------- 
key = 012013012013 and value = 张三 
key = 012013012014 and value = 张四 
 
第二种:通过map.entryset()使用iterator()遍历map2的key和value 
key = 112013012014 and value = 李四 
key = 112013012013 and value = 李三 
map2.size()为:2,b为:2,group2数组的长度为:2 
---------------------------------------------------- 
key = 112013012014 and value = 李四 
key = 112013012013 and value = 李三 
 
第三种:通过map.entryset()遍历map的key和value  
key = 112013012014 and value = 李四 
key = 112013012013 and value = 李三 
key = 012013012013 and value = 张三 
key = 012013012014 and value = 张四 
map.size()为:4,c为:4,group3数组的长度为:4 
---------------------------------------------------- 
key = 112013012014 and value = 李四 
key = 112013012013 and value = 李三 
key = 012013012013 and value = 张三 
key = 012013012014 and value = 张四 

如有疑问请留言或者到本站社区交流讨论,大家共同进步,感谢阅读,希望能帮助到大家,谢谢大家对本站的支持!

如对本文有疑问,请在下面进行留言讨论,广大热心网友会与你互动!! 点击进行留言回复

相关文章:

验证码:
移动技术网