Find the nearest/closest value in a sorted List

Because the collection is sorted, you can do a modified binary search in O( log n ) :

    public static int search(int value, int[] a) {

        if(value < a[0]) {
            return a[0];
        }
        if(value > a[a.length-1]) {
            return a[a.length-1];
        }

        int lo = 0;
        int hi = a.length - 1;

        while (lo <= hi) {
            int mid = (hi + lo) / 2;

            if (value < a[mid]) {
                hi = mid - 1;
            } else if (value > a[mid]) {
                lo = mid + 1;
            } else {
                return a[mid];
            }
        }
        // lo == hi + 1
        return (a[lo] - value) < (value - a[hi]) ? a[lo] : a[hi];
    }

Since most of the code above is binary search, you can leverage the binarySearch(...) provided in the std library and examine the value of the insertion point:

    public static int usingBinarySearch(int value, int[] a) {
        if (value <= a[0]) { return a[0]; }
        if (value >= a[a.length - 1]) { return a[a.length - 1]; }

        int result = Arrays.binarySearch(a, value);
        if (result >= 0) { return a[result]; }

        int insertionPoint = -result - 1;
        return (a[insertionPoint] - value) < (value - a[insertionPoint - 1]) ?
                a[insertionPoint] : a[insertionPoint - 1];
    }

You need Array.binarySearch, docs.

Returns: index of the search key, if it is contained in the array; otherwise, (-(insertion point) - 1). The insertion point is defined as the point at which the key would be inserted into the array: the index of the first element greater than the key, or a.length if all elements in the array are less than the specified key.