如何用对数时间算法解决这个问题？

Question

I came across the following problem:

Now you a given array of double number of length n (n is known) as double sample[n] , and it is in ascending order (all elements are different). Please write a function with one parameter double num , which return the index of the element in sample[n] which is closest to the parameter num . If num is located exactly in the middle of one interval, return the index of the element smaller than it.

Here is my code in Java:

public int getIndex(double num) {
    if(sample[0] >= num) {return 0;}
    for(int i = 1, i < sample.length; i++) {
        if(sample[i] == num) {return i;}
        else if(sample[i] > num) {
            return (sample[i]-num) < (num-sample[i-1]) ? i : i-1;
        else {continue;}
    }
    return sample.length;
}

It is clearly linear complexity. However, I was told by my teacher that algorithm with O(log n) exists. Can anyone help me with the coding?

Answer 1

Your array is sorted, this means you can use binary search . I have nothing to add to this answer and will not provide you ready code, because it is your responsibility. The main idea is that you examine middle element of the array and then determine in which direction you should move: first or second half of an array and so on.