Given an integer array sorted in ascending order, write a function to search target in nums. If target exists, then return its index, otherwise return -1. However, the array size is unknown to you. You may only access the array using an ArrayReader interface, where ArrayReader.get(k) returns the element of the array at index k (0-indexed).

You may assume all integers in the array are less than 10000, and if you access the array out of bounds, ArrayReader.get will return 2147483647.

Example 1:

Input: array = [-1,0,3,5,9,12], target = 9
Output: 4
Explanation: 9 exists in nums and its index is 4

Example 2:

Input: array = [-1,0,3,5,9,12], target = 2
Output: -1
Explanation: 2 does not exist in nums so return -1

Constraints:

You may assume that all elements in the array are unique.
The value of each element in the array will be in the range [-9999, 9999].
The length of the array will be in the range [1, 10^4].

Solution

Time complexity : O(log(n))
Space complexity : O(1)

/**
 * // This is the ArrayReader's API interface.
 * // You should not implement it, or speculate about its implementation
 * class ArrayReader {
 *   public:
 *     int get(int index);
 * };
 */

class Solution {
public:
    int search(const ArrayReader& reader, int target) {
        int l = 0, r = 1;
        while (reader.get(r) < target) {
            l = r;
            r <<= 1;
        }
        printf("r = %d\n", r);
        
        while (l < r) {
            int mid = (l+r)>>1;
            int res = reader.get(mid);
            if (res == target)
                return mid;
            if (target < res)
                r = mid;
            else l = mid+1;
        }
        
        return reader.get(l) == target ? l : -1;
    }
};

流程：找出target範圍 → binary search尋找target。

找出target範圍：window往右移，長度以2的指數成長，直到右界限值大於target。

binary search尋找target：根據上一步找出的範圍，用binary search的方式，找出target。