Search in Unknown Sized Sorted Array

最新推荐文章于 2026-01-01 15:39:09 发布

转载最新推荐文章于 2026-01-01 15:39:09 发布 · 175 阅读

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原文链接：http://www.cnblogs.com/davidnyc/p/8572352.html

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#java

本文介绍了一种针对未知大小且元素按升序排列的整数字典进行目标整数搜索的方法。通过实现特殊的查找算法，该算法首先快速定位到可能的目标范围，再使用二分查找确定具体位置。

Given a integer dictionary A of unknown size, where the numbers in the dictionary are sorted in ascending order, determine if a given target integer T is in the dictionary. Return the index of T in A, return -1 if T is not in A.

Assumptions

dictionary A is not null
dictionary.get(i) will return null(Java)/INT_MIN(C++) if index i is out of bounds

Examples

A = {1, 2, 5, 9, ......}, T = 5, return 2
A = {1, 2, 5, 9, 12, ......}, T = 7, return -1

 1 /*
 2 *  interface Dictionary {
 3 *    public Integer get(int index);
 4 *  }
 5 */
 6 
 7 // You do not need to implement the Dictionary interface.
 8 // You can use it directly, the implementation is provided when testing your solution.
 9 public class Solution {
10   public int search(Dictionary dict, int target) {
11     // Write your solution here
12     //corner case 
13     if(dict == null){
14         return -1 ; 
15     }
16        //corner case: index 0 == target
17     if(dict.get(0) != null && dict.get(0) == target){
18         return 0 ; 
19     }
20     int left = 1 ; 
21     while(dict.get(left) != null && dict.get(left)<target){
22         left *= 2 ; 
23     }
24     int start = left/2 , end = left ; 
25     while(start +1 < end){
26         int mid = start + (end - start)/2 ; 
27          if(dict.get(mid) != null && dict.get(mid) == target){
28           return mid ; 
29       } else if(dict.get(mid)!= null && dict.get(mid) < target){
30           start = mid ; 
31       } else{
32           end = mid ; 
33       }
34     }
35     //post processing
36     if(dict.get(start)!= null && dict.get(start) == target){
37         return start ; 
38     } 
39     if(dict.get(end)!= null && dict.get(end) == target){
40         return end ; 
41     }
42     return -1;
43   }
44 }