一个简单的Java集合范围过滤的多个方式对比

在一个项目里面有这么一个技术需求:
1.集合中元素个数,10M
2.根据上限和下限从一个Set中过滤出满足要求的元素集合.

实际这个是个很典型的技术要求, 之前的项目也遇见过,但是因为当时的类库不多, 都是直接手写实现的. 方式基本等同于第一个方式.

在这个过程中, 我写了四个方式, 基本记录到下面.
第一个方式:对Set进行迭代器遍历, 判断每个元素是否都在上限和下限范围中.如果满足则添加到结果集合中, 最后返回结果集合.
            测试效果:集合大小100K, 运算时间 3000ms+
过滤部分的逻辑如下:

 1     void filerSet(Set<BigDecimal> targetSet, String lower, String higher) {
 2         BigDecimal bdLower = new BigDecimal(Double.parseDouble(lower));
 3         BigDecimal bdHigher = new BigDecimal(Double.parseDouble(higher));
 4 
 5         Set<BigDecimal> returnSet = new HashSet<BigDecimal>();
 6         for (BigDecimal object : targetSet) {
 7             if (isInRange(object, bdLower, bdHigher)) {
 8                 returnSet.add(object);
 9             }
10         }
11     }
12 
13     private boolean isInRange(BigDecimal object, BigDecimal bdLower,
14             BigDecimal bdHigher) {
15         return object.compareTo(bdLower) >= 0
16                 && object.compareTo(bdHigher) <= 0;
17     }

第二个方式: 借助TreeSet, 原始集合进行排序, 然后直接subset.
            测试效果: 集合大小10M, 运算时间: 12000ms+(获得TreeSet) , 200ms(获得结果)
过滤部分的逻辑如下(非常繁琐):

  1     Set<BigDecimal> getSubSet(TreeSet<BigDecimal> targetSet, String lower,
  2             String higher) {
  3 
  4         BigDecimal bdLower = new BigDecimal(Double.parseDouble(lower));
  5         BigDecimal bdHigher = new BigDecimal(Double.parseDouble(higher));
  6 
  7         if ((bdHigher.compareTo(targetSet.first()) == -1)
  8                 || (bdLower.compareTo(targetSet.last()) == 1)) {
  9             return null;
 10         }
 11 
 12         boolean hasLower = targetSet.contains(bdLower);
 13         boolean hasHigher = targetSet.contains(bdHigher);
 14         if (hasLower) {
 15             if (hasHigher) {
 16                 System.out.println("get start:" + bdLower);
 17                 System.out.println("get end:" + bdHigher);
 18                 return targetSet.subSet(bdLower, true, bdHigher, true);
 19             } else {
 20                 BigDecimal newEnd = null;
 21                 System.out.println("get start:" + bdLower);
 22                 SortedSet<BigDecimal> returnSet = null;
 23                 if (bdHigher.compareTo(targetSet.last()) != -1) {
 24                     newEnd = targetSet.last();
 25                 } else {
 26                     SortedSet<BigDecimal> newTargetSet = targetSet
 27                             .tailSet(bdLower);
 28                     for (BigDecimal object : newTargetSet) {
 29                         if (object.compareTo(bdHigher) == 1) {
 30                             newEnd = object;
 31                             break;
 32                         } else if (object.compareTo(bdHigher) == 0) {
 33                             newEnd = object;
 34                             break;
 35                         }
 36                     }
 37                 }
 38                 returnSet = targetSet.subSet(bdLower, true, newEnd, true);
 39                 if (newEnd.compareTo(bdHigher) == 1) {
 40                     returnSet.remove(newEnd);
 41                 }
 42                 return returnSet;
 43             }
 44 
 45         } else {
 46             if (hasHigher) {
 47                 System.out.println("get end:" + bdHigher);
 48                 TreeSet<BigDecimal> newTargetSet = (TreeSet<BigDecimal>) targetSet
 49                         .headSet(bdHigher, true);
 50                 BigDecimal newStart = null;
 51                 SortedSet<BigDecimal> returnSet = null;
 52 
 53                 if (bdLower.compareTo(targetSet.first()) == -1) {
 54                     newStart = targetSet.first();
 55                 } else {
 56                     for (BigDecimal object : newTargetSet) {
 57                         if (object.compareTo(bdLower) != -1) {
 58                             newStart = object;
 59                             break;
 60                         }
 61                     }
 62                 }
 63                 returnSet = targetSet.subSet(newStart, true, bdHigher, true);
 64 
 65                 return returnSet;
 66             } else {
 67                 System.out.println("Not get start:" + bdLower);
 68                 System.out.println("Not get end:" + bdHigher);
 69                 BigDecimal newStart = null;
 70                 BigDecimal newEnd = null;
 71                 if (bdHigher.compareTo(targetSet.last()) != -1) {
 72                     newEnd = targetSet.last();
 73                 }
 74                 if (bdLower.compareTo(targetSet.first()) == -1) {
 75                     newStart = targetSet.first();
 76                 }
 77                 for (BigDecimal object : targetSet) {
 78                     if (newStart == null) {
 79                         if (object.compareTo(bdLower) != -1) {
 80                             newStart = object;
 81                             if (newEnd != null) {
 82                                 break;
 83                             }
 84                         }
 85                     }
 86 
 87                     if (newEnd == null) {
 88                         if (object.compareTo(bdHigher) != -1) {
 89                             newEnd = object;
 90                             if (newStart != null) {
 91                                 break;
 92                             }
 93                         }
 94                     }
 95                 }
 96 
 97                 if (newStart == null) {
 98                     if (newEnd == null) {
 99                         if ((bdHigher.compareTo(targetSet.first()) == -1)
100                                 || (bdLower.compareTo(targetSet.last()) == 1)) {
101                             return null;
102                         }
103                         return targetSet;
104                     } else {
105                         SortedSet<BigDecimal> newTargetSet = targetSet.headSet(
106                                 newEnd, true);
107                         if (newEnd.compareTo(bdHigher) == 1) {
108                             newTargetSet.remove(newEnd);
109                         }
110                         return newTargetSet;
111                     }
112                 } else {
113                     if (newEnd == null) {
114                         SortedSet<BigDecimal> newTargetSet = targetSet.tailSet(
115                                 newStart, true);
116                         return newTargetSet;
117                     } else {
118                         SortedSet<BigDecimal> newTargetSet = targetSet.subSet(
119                                 newStart, true, newEnd, true);
120                         if (newEnd.compareTo(bdHigher) == 1) {
121                             newTargetSet.remove(newEnd);
122                         }
123                         return newTargetSet;
124                     }
125                 }
126             }
127         }
128     }

第三种方式: 使用Apache Commons Collections, 直接对于原始Set进行filter.
            测试效果:集合大小10M,过滤结果1M, 运算时间: 1000ms+
过滤部分的代码如下:

 1 //过滤的主体逻辑
 2     void filterSet(Set<BigDecimal> targetSet, String lower, String higher) {
 3         final BigDecimal bdLower = new BigDecimal(Double.parseDouble(lower));
 4         final BigDecimal bdHigher = new BigDecimal(Double.parseDouble(higher));
 5 
 6         Predicate predicate = new Predicate() {
 7             public boolean evaluate(Object object) {
 8                 BigDecimal bDObject = (BigDecimal) object;
 9                 return bDObject.compareTo(bdLower) >= 0
10                         && bDObject.compareTo(bdHigher) <= 0;
11             }
12         };
13 
14         CollectionUtils.filter(targetSet, predicate);

15　　　/*　CollectionUtils.filter(targetSet, new Predicate() {
16　　　　@Override
17　　　　public boolean evaluate(Object o) {
18　　　　BigDecimal bDObject = (BigDecimal) o;
19　　　　 return bDObject.compareTo(bdLower) >= 0
10                         && bDObject.compareTo(bdHigher) <= 0;
20　　　　}
21　　　　});*/
22     }

第四种方式:使用Guava(google Collections), 直接对于原始Set进行Filter
            测试效果:集合大小10M,过滤结果1M, 运算时间: 100ms-
过滤部分的代码如下:

 1 //guava filter
 2 
 3     Set<BigDecimal> filterSet(Set<BigDecimal> targetSet, String lower,
 4             String higher) {
 5         final BigDecimal bdLower = new BigDecimal(Double.parseDouble(lower));
 6         final BigDecimal bdHigher = new BigDecimal(Double.parseDouble(higher));
 7 
 8         Set<BigDecimal> filterCollection = Sets.filter(targetSet,
 9                 new Predicate<BigDecimal>() {
10                     @Override
11                     public boolean apply(BigDecimal input) {
12                         BigDecimal bDObject = (BigDecimal) input;
13                         return bDObject.compareTo(bdLower) >= 0
14                                 && bDObject.compareTo(bdHigher) <= 0;
15                     }
16                 });
17 
18         return filterCollection;
19     }

四种方式对比如下:
第一种方式:  仅依赖于JAVA原生类库 遍历时间最慢, 代码量很小
第二种方式:  仅依赖于JAVA原生类库 遍历时间比较慢(主要慢在生成有序Set), 代码量最多
第三种方式:  依赖于Apache Commons Collections, 遍历时间比较快, 代码量很少
第四种方式:  依赖于Guava, 遍历时间最快, 代码量很少

基于目前个人的技术水平和视野, 第四种方式可能是最佳选择.

记录一下, 以后可能还会有更好的方案.