LeetCode 702. Search in a Sorted Array of Unknown Size
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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 innums
and its index is 4
Example 2:
Input:array
= [-1,0,3,5,9,12],target
= 2 Output: -1 Explanation: 2 does not exist innums
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 }
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