原题链接在这里：https://leetcode.com/problems/search-in-a-sorted-array-of-unknown-size/
题目：
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
Note:
- 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].
题解：
The range of index could not be over 20000. Because element raget is [-9999, 9999].
Guess the index using binary search, and call the reader.get() on the guess index.
If the return value cur > target, it could be either there is no such index, return is Integer.MAX_VALUE, or it exists, but value is larger. Either way, should continue guessing smaller index.
If cur < target, should continue guessing larger index.
If cur == target, return the guess index.
Time Complexity: O(logn). n = 20000.
Space:O(1).
AC Java:
1 class Solution {
2 public int search(ArrayReader reader, int target) {
3 int l = 0;
4 int r = 20000;
5 while(l <= r){
6 int mid = l + (r-l)/2;
7 int cur = reader.get(mid);
8 if(cur > target){
9 r = mid-1;
10 }else if(cur < target){
11 l = mid+1;
12 }else{
13 return mid;
14 }
15 }
16
17 return -1;
18 }
19 }
Could use candidate call to get r. while (reader.get(r) < target). r *=2.
Then target index should be (r/2,r].
Time Complexity: O(logn).
Space: O(1).
AC Java:
1 class Solution {
2 public int search(ArrayReader reader, int target) {
3 int r = 1;
4 while(reader.get(r) < target){
5 r = r << 1;
6 }
7
8 int l = r >> 1;
9 while(l <= r){
10 int mid = l + (r-l)/2;
11 int cur = reader.get(mid);
12 if(cur > target){
13 r = mid-1;
14 }else if(cur < target){
15 l = mid+1;
16 }else{
17 return mid;
18 }
19 }
20
21 return -1;
22 }
23 }