检查数组中所有元素是否相等的最快方法

Question

What would be the fastest way preferably native to java to check if all elements of an array are equal to a value? (denoted here by n)

So far I have:

boolean check = true;
int len = times.length;
for(int a = 0; check && a < len; a++) {
    check = times[a]==n && check;
}

So if each element is equal to a value, check is set to true, otherwise it is set to false.

EDIT: Would this be faster?

boolean check = true;
int len = times.length
int a = 0;
while(a < len && times[a]==n) {
    a++;
}
check=(a==len);

Ok, after looking at the answers here I understand the code is as small as its gonna get, so I'll have to look into threads and parallel processing, thanks everyone for the help and the links

Answer 1

In Java 8, you could use the Stream API :

boolean check = Arrays.asList(times).stream().allMatch(t -> t == n);

This alone won't be faster than iterating over the array directly. But if you then switch to a parallel stream , it could be significantly faster on large arrays. And if performance is a concern, it's presumably large arrays that you care about.

boolean check = Arrays.asList(times).parallelStream().allMatch(t -> t == n);

A parallel stream allows the array scanning to be parcelled out to multiple threads, scanning the array using parallel CPUs or cores.

Answer 2

This algorithm is O(n) which is the fastest way to check all elements of a list because you only have to check each element only one time.

Now just because this is the fastest algorithm in finding if all elements equal a value doesn't mean you have optimized it to its fullest potential.

This then leaves room for a multi-threaded/multiprocessor implementation.

A solution using more cores or threads is splitting the array into the amount of thread/cores you want to process simultaneously ie if you have an array of 100 elements and want 10 threads running at the same time-- split the array into 10 pieces and run each section on the array.

Some psudo code:

 int from,to, threadCount = 10;
 boolean check[threadCount];  

 int factor = array.length()/threadCount;
 for(int i = 0; i < threadCount; i++){
      from = i*factor;
      to = i*factor+factor;
      newThreadProcess(from, to, array, check[i]);     
 }

 barrier(); //Wait till all the threads are done processing
 for(int i = 0; i < threadCount; i++){
    if(!check[i]) return false;
 }
 return true;

This is best when the array is very large

Answer 3

check = times[a]==n && check;

can be

check = times[a]==n;

since you never would've gotten to that line unless check were true.

Answer 4

A simple and clean way:

public static boolean allEqual(E[] array, E element) {
    for(E e : array) {
        if(e != element) {
            return false;
        }
    }
    return true;
}

Answer 5

你可以用更短的方式写它：

for(int a = 0; a < len && (check = times[a]==n); a++);