LeetCode 702. Search in a Sorted Array of Unknown Size

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:

1. You may assume that all elements in the array are unique.
2. 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:

 class Solution {
     public int search(ArrayReader reader, int target) {
         int l = 0;
         int r = 20000;
         while(l <= r){
             int mid = l + (r-l)/2;
             int cur = reader.get(mid);
             if(cur > target){
                 r = mid-1;
             }else if(cur < target){
                 l = mid+1;
             }else{
                 return mid;
             }
         }

         return -1;
     }
 }

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:

 class Solution {
     public int search(ArrayReader reader, int target) {
         int r = 1;
         while(reader.get(r) < target){
             r = r << 1;
         }

         int l = r >> 1;
         while(l <= r){
             int mid = l + (r-l)/2;
             int cur = reader.get(mid);
             if(cur > target){
                 r = mid-1;
             }else if(cur < target){
                 l = mid+1;
             }else{
                 return mid;
             }
         }

         return -1;
     }
 }

类似Binary Search.