Given a big sorted array with positive integers sorted by ascending order. The array is so big so that you can not get the length of the whole array directly, and you can only access the kth number by ArrayReader.get(k) (or ArrayReader->get(k) for C++). Find the first index of a target number. Your algorithm should be in O(log k), where k is the first index of the target number.
Return -1, if the number doesn't exist in the array.
Example
Given [1, 3, 6, 9, 21, ...], and target = 3, return 1.
Given [1, 3, 6, 9, 21, ...], and target = 4, return -1.
分析：
给一个按照升序排序的正整数数组。这个数组很大以至于你只能通过固定的接口ArrayReader.get(k) 来访问第k个数。(或者C++里是ArrayReader->get(k))，并且你也没有办法得知这个数组有多大。找到给出的整数target第一次出现的位置。你的算法需要在O(logk)的时间复杂度内完成，k为target第一次出现的位置的下标。
如果找不到target，返回-1。
比如 nums数组：[ 1 4 5 7 8 10 15 20 30 50 80 90 100 200 300 400 ….]，
这个很大很大的数组，要找20，由于数组很大，所以要先找到nums[index]的值大于20的，最快速的方法是找到nums(index)大于20的，
依次找nums[2], nums[4]，nums[8]，找到第一个大于20的数，为nums[8], 所以循环要从nums[0]作为起点，nums[8]作为终点.
/**
* Definition of ArrayReader:
*
* class ArrayReader {
* // get the number at index, return -1 if index is less than zero.
* public int get(int index);
* }
*/
public class Solution {
/**
* @param reader: An instance of ArrayReader.
* @param target: An integer
* @return : An integer which is the index of the target number
*/
public int searchBigSortedArray(ArrayReader reader, int target) {
int index = 1;
while (reader.get(index - 1) < target) {
index = index * 2;
}
int start = 0, end = index - 1;
while (start + 1 < end) {
int mid = start + (end - start) / 2;
if (reader.get(mid) < target) {
start = mid;
} else {
end = mid;
}
} if (reader.get(start) == target) {
return start;
} if (reader.get(end) == target) {
return end;
}
return -1;
}
}