java determines whether the array elements are repeated:
1. First, deduplicate the array, and then determine whether the array elements are repeated by judging the array length.
/**
* 判斷數(shù)組內(nèi)有無重復(fù)元素
* @param args
* @return true 有重復(fù) | false 無重復(fù)
*/
public static boolean hasRepeat(Object[] args){
Set<Object> tempSet = new HashSet<Object>();
for (int i = 0; i < args.length; i++) {
tempSet.add(args[i]);
}
if(args.length == tempSet.size()){
return false;
}else{
return true;
}
}2. Compare the elements in the array for equality through double loops to determine whether the elements are repeated
1. Two for loops, determine whether they are equal in sequence, or call Arrays.sort() first, Then use Arrays.binarysearch() to compare
2, convert the array into a string separated by special characters, and then use \1 in the regular expression, which is the first matching result. This method Higher efficiency.
public void testMethod(){
int []b=new int[]{1,3,5,6,2,4,20,9};
boolean flag=true;
for(int i=0;i<b.length-1;i++){
for(int j=i+1;j<b.length;j++){ //注意FOR嵌套的用法
if(b[i]==b[j]){
flag=false;
//break;
}else{
flag=true;
//break;
}
}
}
if(flag){
System.out.println("不重復(fù)");
}else{
System.out.println("重復(fù)");
}
}For more java knowledge, please pay attention to java basic tutorial.
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