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算法:支持重复元素的二分查找

时间:2015-01-09 00:12:37      阅读:361      评论:0      收藏:0      [点我收藏+]

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近几天在处理的一个项目,需要频繁对一些有序超大集合进行目标查找,二分查找算法是这类问题的最优解。但是java的Arrays.binarySearch()方法,如果集合中有重复元素,而且遇到目标元素正好是这些重复元素之一,该方法只能返回一个,并不能将所有的重复目标元素都返回,没办法,只能自造轮子了。

先复习下二分查找的经典算法:

 1     private int binarySearch1(Integer[] A, Integer x) {
 2         int low = 0, high = A.length - 1;
 3         while (low <= high) {
 4             int mid = (low + high) / 2;
 5             if (A[mid].equals(x)) {
 6                 return mid;
 7             } else if (x < A[mid]) {
 8                 high = mid - 1;
 9             } else {
10                 low = mid + 1;
11             }
12         }
13         return -1;
14     }

思路很简单,先定位到中间元素,如果中间元素比目标元素大,则扔掉后一半,反之扔掉前一半,如果正好一次命中,直接返回。

略做改进:

 1     private List<Integer> binarySearch2(Integer[] A, Integer x) {
 2         List<Integer> result = new ArrayList<Integer>();
 3         int low = 0, high = A.length - 1;
 4         while (low <= high) {
 5             int mid = (low + high) / 2;
 6             if (A[mid].equals(x)) {
 7                 if (mid > 0) {
 8                     //看前一个元素是否=目标元素
 9                     if (A[mid - 1].equals(x)) {
10                         for (int i = mid - 1; i >= 0; i--) {
11                             if (A[i].equals(x)) {
12                                 result.add(i);
13                             } else break;
14                         }
15                     }
16                 }
17                 result.add(x);
18                 if (mid < high) {
19                     //看后一个元素是否=目标元素
20                     if (A[mid + 1].equals(x)) {
21                         for (int i = mid + 1; i <= high; i++) {
22                             if (A[i].equals(x)) {
23                                 result.add(i);
24                             } else break;
25                         }
26                     }
27                 }
28                 return result;
29             } else if (x < A[mid]) {
30                 high = mid - 1;
31             } else {
32                 low = mid + 1;
33             }
34         }
35         return result;
36 
37     }

思路:命中目标后,看下前一个紧挨着的元素是否也是要找的元素,如果是,则顺着向前取,直到遇到不等于目标元素为止。然后再看后一个紧挨着的元素,做类似处理。

测试:

1         Integer[] A = new Integer[]{1, 2, 3, 4, 5, 5, 5, 6, 7, 8, 9};
2 
3         System.out.println("binarySearch1 => ");
4         System.out.println(binarySearch1(A, 5));
5 
6         System.out.println("binarySearch2 => ");
7         System.out.println(binarySearch2(A, 5));

binarySearch1 =>
5
binarySearch2 =>
[4, 5, 6]

从返回的下标值看,都在预期之中,但是事情并未到此止步,通常要查找的列表元素,并不是数值这么简单,一般是一些复杂的对象实例,为了做到通用,得弄成一个泛型版本:

 1    private <T> List<Integer> binarySearch4(List<T> A, T x, Comparator<? super T> comparator) {
 2         List<Integer> result = new ArrayList<Integer>();
 3         int low = 0, high = A.size() - 1;
 4         while (low <= high) {
 5             int mid = (low + high) / 2;
 6             int temp = comparator.compare(x, A.get(mid));
 7             if (temp == 0) {
 8                 if (mid > 0) {
 9                     if (comparator.compare(x, A.get(mid - 1)) == 0) {
10                         for (int i = mid - 1; i >= 0; i--) {
11                             if (comparator.compare(A.get(i), x) == 0) {
12                                 result.add(i);
13                             } else break;
14                         }
15                     }
16                 }
17                 result.add(mid);
18                 if (mid < high) {
19                     if (comparator.compare(x, A.get(mid + 1)) == 0) {
20                         for (int i = mid + 1; i <= high; i++) {
21                             if (comparator.compare(x, A.get(i)) == 0) {
22                                 result.add(i);
23                             } else break;
24                         }
25                     }
26                 }
27                 return result;
28 
29             } else if (temp < 0) {
30                 high = mid - 1;
31             } else {
32                 low = mid + 1;
33             }
34         }
35 
36         return result;
37     }

为了比较二个复杂对象实例的大小,引入了Comparator接口,可以根据业务需要,则调用者自定义比较规则。

测试一下:

先定义一个业务对象类AwbDto:

 1 package com.cnblogs.yjmyzz.test.domain;
 2 
 3 /**
 4  * Created by jimmy on 15/1/8.
 5  */
 6 public class AwbDto {
 7 
 8 
 9     /**
10      * 运单号
11      */
12     private String awbNumber;
13 
14     /**
15      * 始发站
16      */
17     private String originAirport;
18 
19     public AwbDto(String awbNumber, String originAirport) {
20         this.awbNumber = awbNumber;
21         this.originAirport = originAirport;
22     }
23 
24     public String getAwbNumber() {
25         return awbNumber;
26     }
27 
28     public void setAwbNumber(String awbNumber) {
29         this.awbNumber = awbNumber;
30     }
31 
32     public String getOriginAirport() {
33         return originAirport;
34     }
35 
36     public void setOriginAirport(String originAirport) {
37         this.originAirport = originAirport;
38     }
39 }

还需要定义AwbData比较大小的业务规则,假设:只要运单号相同,则认为相等(即:用运单号来区分对象大小)

1     private class AwbDtoComparator implements Comparator<AwbDto> {
2 
3         @Override
4         public int compare(AwbDto x, AwbDto y) {
5             return x.getAwbNumber().compareTo(y.getAwbNumber());
6         }
7     }

测试代码:

 1         List<AwbDto> awbList = new ArrayList<AwbDto>();
 2         awbList.add(new AwbDto("112-10010011", "北京"));
 3         awbList.add(new AwbDto("112-10010022", "上海"));
 4         awbList.add(new AwbDto("112-10010033", "天津"));
 5         awbList.add(new AwbDto("112-10010044", "武汉"));
 6         awbList.add(new AwbDto("112-10010044", "武汉"));
 7         awbList.add(new AwbDto("112-10010055", "广州"));
 8 
 9         AwbDtoComparator comparator = new AwbDtoComparator();
10 
11         AwbDto x = new AwbDto("112-10010044", "武汉");
12 
13 
14         System.out.println("binarySearch4 => ");
15         System.out.println(binarySearch4(awbList, x, comparator));

binarySearch4 =>
[3, 4]

测试结果,一切顺利,皆大欢喜,可以去休息了。

算法:支持重复元素的二分查找

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原文地址:http://www.cnblogs.com/yjmyzz/p/4212149.html

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