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.
Notice
If you accessed an inaccessible index (outside of the array), ArrayReader.get will return 2,147,483,647.
Example
Given [1, 3, 6, 9, 21, ...], and target = 3, return 1.
Given [1, 3, 6, 9, 21, ...], and target = 4, return -1.
Challenge
O(log k), k is the first index of the given target number.
难点在于end不能直接取到，所以一开始二倍增长地去找一个可能性end（target <= reader.get(end)）。接下来就是普通的二分查找了。
public class Solution {
/*
* @param reader: An instance of ArrayReader.
* @param target: An integer
* @return: An integer which is the first index of target.
*/
public int searchBigSortedArray(ArrayReader reader, int target) {
final int MAX = 2147483647;
int start = 0;
int end = 1;
int flip = 1;
while (reader.get(end) < target){
end += flip;
flip *= 2;
if (flip == 2 && reader.get(end) == MAX){
return -1;
}
if (reader.get(end) == MAX){
flip /= 2;
end -= flip;
flip = 1;
}
}
while (start + 1 < end){
int mid = start + (end - start) / 2;
if (target < reader.get(mid)){
end = mid;
} else if (target == reader.get(mid)){
end = mid;
} else {
start = mid;
}
}
if (target == reader.get(start)){
return start;
}
if (target == reader.get(end)){
return end;
}
return -1;
}
}