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.
class Solution {
public:
/**
* @param reader: An instance of ArrayReader.
* @param target: An integer
* @return: An integer which is the first index of target.
*/
int searchBigSortedArray(ArrayReader *reader, int target) {
// write your code here
int index = 0;
while (reader->get(index) != -1 && reader->get(index) < target ) {
index = index * 2 + 1;
}
int start = 0;
int end = index;
while (start + 1 < end) {
int mid = start + (end - start) / 2;
if (reader->get(mid) == target) {
end = mid;
} else if (reader->get(mid) == -1 || reader->get(mid) > target) {
end = mid;
} else {
start = mid;
}
}
if (reader->get(start) == target) {
return start;
}
if (reader->get(end) == target) {
return end;
}
return -1;
}
};
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) {
// write your code here
if (reader == null) {
return -1;
}
int start = 0;
int end = Integer.MAX_VALUE;
int res = -1;
while (start <= end) {
int mid = start + (end - start) / 2;
if (reader.get(mid) == -1) {
end = mid - 1;
continue;
}
if (reader.get(mid) == target) {
res = mid;
end = mid - 1;
} else if (reader.get(mid) < target) {
start = mid + 1;
} else {
end = mid - 1;
}
}
return res;
}
}