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。