https://blog.csdn.net/love_java_cc/article/details/56674401

首先，看一下源码：

public V put(K paramK, V paramV)  {if (paramK == null) {return putForNullKey(paramV);    }int i = hash(paramK.hashCode());int j = indexFor(i, this.table.length);for (Entry localEntry = this.table[j]; localEntry != null; localEntry = localEntry.next)    {      Object localObject1;if ((localEntry.hash == i) && (((localObject1 = localEntry.key) == paramK) || (paramK.equals(localObject1))))      {Object localObject2 = localEntry.value;        localEntry.value = paramV;        localEntry.recordAccess(this);return localObject2;      }    }this.modCount += 1;    addEntry(i, paramK, paramV, j);return null;  }

好吧，看不懂没关系，下面看例子：

import java.util.HashMap;import java.util.Map;public class Test {public static void main(String[] args) {		Map<String, String> map = new HashMap<String, String>();String p1 = map.put("11", "22");		System.out.println("p1:" + p1);String p2 = map.put("33", "44");		System.out.println("p2:" + p2);String value1 = map.get("11");		System.out.println("value1:" + value1);String p3 = map.put("11", "44");		System.out.println("p3：" + p3);String value2 = map.get("11");		System.out.println("value2:" + value2);	}}

输出结果：

p1:nullp2:nullvalue1:22p3：22value2:44

说明：put方法返回值为null或者value；

如果key没有重复，put成功，则返回null，如p1、p2；

如果key重复了，返回的是map.get(key)，也就是当前这个key对应的value，如上面的p3，key="11"，而p1的key也是"11"，p1与p3重复，返回的是p1的value="22"，并且将p3覆盖掉p1