How to check if an array of numbers has gaps?

问题

I have a Long array with these numbers:

long[] = {1,2,3,5,6,7};

Notice that 4 is missing. What's the best way to test this array if any such gaps exist or not?

回答1:

If you're guaranteed that arrays is ordered without any duplicate then you could check that in O(1)

I think this code should work in this specific case :)

//assume that given array is ordered and has no duplicated value
long[] myarray = {5,6,7};                 //no gap
long[] myarray1 = {1,2,4};                //has gap
long[] myarray2 = {10,11,12,13,14,15};    //no gap

//return true if has gap
//return false if no gap
//throw null-pointer if empty
public static boolean checkIfHasGap(long[] array) {
    if (array.length == 0) {
        throw new NullPointerException("Given Array is empty");
    } else {
        return array[0] + array.length != array[array.length - 1] + 1;
    }
}

回答2:

public static boolean hasGaps(long[] array) {
    if (array == null || array.length == 0) {
        return false;
    }
    if (array[array.length - 1] - array[0] + 1 != array.length) {
        return true;
    }
    for (int i = 1; i < array.length; i++) {
        if (array[i] != array[i - 1] + 1) {
            return true;
        }
    }
    return false;
}

This method first check for the "easy" cases, then iterate the array to check the more complex cases.

回答3:

Loop through the array and check, if the current item is exactly x more than the current index, where x is the first element in your array.

public boolean hasGaps(long[] array) {
    if(array == null || array.length == 0) {
        return false;
    }

    long start = array[0];
    for(int i = 0; i < array.length; i++) {
        if(array[i] != i + start) {
            return true;
        }
    }
    return false;
}

来源：https://stackoverflow.com/questions/30053354/how-to-check-if-an-array-of-numbers-has-gaps