Java如何将数组中具有相同的元素都删去 Java数组中重复数据的删除
java\u4e2d\u600e\u4e48\u5c06\u5b57\u7b26\u4e32\u6570\u7ec4\u4e2d\u7684\u91cd\u590d\u7684\u5143\u7d20\u53bb\u6389\uff0c\u4f46\u662f\u8fd8\u662f\u4fdd\u7559\u4e00\u4e2a\u3002import java.util.*;
class BaiDu
{
public static void main(String[] args)
{
TreeSet tr = new TreeSet();
String[] s ={"11","22","22","33","33","33"};
System.out.print("====\u5904\u7406\u524d=======");
for(int i=0;i<s.length;i++){
System.out.print(s[i]+" ");
tr.add(s[i]);
}
String[] s2= new String[tr.size()];
System.out.println("=====\u5904\u7406\u540e======");
for(int i=0;i<s2.length;i++){
s2[i]=tr.pollFirst();//\u4eceTreeSet\u4e2d\u53d6\u51fa\u5143\u7d20\u91cd\u65b0\u8d4b\u7ed9\u6570\u7ec4
System.out.print(s2[i]+" ");
}
}
}
java\u6570\u7ec4\u4e2d\u53bb\u6389\u91cd\u590d\u6570\u636e\u53ef\u4ee5\u4f7f\u7528set\u96c6\u5408\uff0cset\u672c\u8eab\u5185\u90e8\u662f\u4e0d\u4fdd\u5b58\u91cd\u590d\u7684\u6570\u636e\u7684\uff0c\u5982\u4e0b\u4ee3\u7801\uff1a
import java.util.Arrays;import java.util.Set;import java.util.TreeSet;public class ceshi {public static void main(String[] args) {int[] testArr = { 5, 5, 1, 2, 3, 6, -7, 8, 6, 45, 5 };//\u65b0\u5efa\u4e00\u4e2aint\u7c7b\u578b\u6570\u7ec4System.out.println(Arrays.toString(testArr));Set set = new TreeSet();//\u65b0\u5efa\u4e00\u4e2aset\u96c6\u5408for (int i : testArr) {set.add(i);}Integer[] arr2 = set.toArray(new Integer[0]);// \u6570\u7ec4\u7684\u5305\u88c5\u7c7b\u578b\u4e0d\u80fd\u8f6c \u53ea\u80fd\u81ea\u5df1\u8f6c\uff1b\u5427Integer\u8f6c\u4e3a\u4e3aint\u6570\u7ec4\uff1bint[] result = new int[arr2.length];for (int i = 0; i < result.length; i++) {result[i] = arr2[i];}System.out.println(Arrays.toString(arr2));}}\u8fd0\u884c\u7ed3\u679c\u5982\u4e0b\uff1a
比如现在有个数组为 myArray ，里面有部分的重复元素
Set mySet = new HashSet();
for(Object obj : Array){
mySet.add(obj);
}
mySet中所保存的元素就是唯一的了.
再吧mySet保存到数组中
完整例子：
// 创建一个数组，里面存在重复的元素
String[] myArray = {"s","s","f","d"};
Set<String> mySet = new HashSet<String>();
// 去除重复元素
for(String s : myArray){
mySet.add(s);
}
myArray = new String[mySet.size()];
int index = 0;
// 将去重后的结果存入数组
for(String s : mySet){
myArray[index] = s;
index++;
}
// 打印出来结果
System.out.println(Arrays.toString(myArray));
ArrayList a = new ArrayList(数组);
for(int i = 0 ;i<a.size();i++){
Object obj = 数组[i];
if(a.contains(obj)){
a.remove(i);
}
}
我答案错了，不好意思，当初咋想的忘记了，抱歉。下面的其他匿名网友回答是正确的。另外附上我的修改。
一：利用set的唯一性解决
String[] strArr = {"B","A","C","D","C","A"};
Set<String> strList = new HashSet(Arrays.asList(strArr));
System.out.println(strList.toString());二：创建新的存储单位，判断是否重复后逐个增加
public static void main(String[] args){
String[] strArr = {"B","A","C","D","C","A"};
List<String> strList = new ArrayList<String>();
for (String str : strArr){
if(strList.contains(str)){
continue;
}
strList.add(str);
}
System.out.println(strList.toString());
}三：利用steam的distinct方法去重(jdk1.8的新特性)
String[] strArr = {"B","A","C","D","C","A"};
List<String> strList = Arrays.asList(strArr);
strList = strList.stream().distinct().collect(Collectors.toList());
System.out.println(strList.toString());
用HashSet，将所有的元素放入到hashset里面，再读出来相同的元素就都删除只剩下一个了。
将数组放入set中后在转化到数组就行啦。
想知道，你最后是如何解决的
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