java排序

转x33g5p2x  于2022-02-09 转载在 Java  
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java排序

Map根据key排序

  1. //升序
  2. Map<Integer,String> treeMap = new TreeMap<>();
  3.         treeMap.put(1,"111");
  4.         treeMap.put(8,"888");
  5.         treeMap.put(4,"444");
  6.         treeMap.put(2,"222");
  7. System.out.println(treeMap);  //输出{1=111, 2=222, 4=444, 8=888}
  9. //降序：
  10.  Map<Integer,String> treeMap = new TreeMap<>((o1,o2)->{
  11.             return o2.compareTo(o1);
  12.         });
  13.         treeMap.put(1,"111");
  14.         treeMap.put(8,"888");
  15.         treeMap.put(4,"444");
  16.         treeMap.put(2,"222");
  17.  System.out.println(treeMap);输出{8=888, 4=444, 2=222, 1=111}

Map根据value排序

对value降序：

  1. Map<String, Double> budget = new HashMap<>();
  2.         budget.put("普通账户", 1000.00);
  3.         budget.put("信用账户", 2000.00);
  4.         budget.put("期权账户",500.00);
  5.         budget.put("理财账户", 4000.00);
  6. //jdk8通过stream实现对Map降序排序
  7.    Map<String, Double> sorted = budget
  8.                 .entrySet()
  9.                 .stream()
  10.                 .sorted(Map.Entry.comparingByValue(Comparator.reverseOrder()))
  11.                 .collect(Collectors.toMap(Map.Entry::getKey, Map.Entry::getValue, (e1, e2) -> e2, LinkedHashMap::new));
  12. //sorted为{理财账户=4000.0, 信用账户=2000.0, 普通账户=1000.0, 期权账户=500.0}

对value降序并只要前俩个：

  1. Map<String, Double> budget = new HashMap<>();
  2.         budget.put("普通账户", 1000.00);
  3.         budget.put("信用账户", 2000.00);
  4.         budget.put("期权账户",500.00);
  5.         budget.put("理财账户", 4000.00);
  7. //jdk8通过stream实现对Map降序排序
  8.    Map<String, Double> sorted = budget
  9.                 .entrySet()
  10.                 .stream()
  11.                 .sorted(Map.Entry.comparingByValue(Comparator.reverseOrder()))
  12.                 .limit(2)
  13.                 .collect(Collectors.toMap(Map.Entry::getKey, Map.Entry::getValue, (e1, e2) -> e2, LinkedHashMap::new));
  14. //sorted为{理财账户=4000.0, 信用账户=2000.0}

对value升序：

  1. Map<String, Double> budget = new HashMap<>();
  2.         budget.put("普通账户", 1000.00);
  3.         budget.put("信用账户", 2000.00);
  4.         budget.put("期权账户",500.00);
  5.         budget.put("理财账户", 4000.00);
  7. //jdk8通过stream实现对Map升序排序
  8.         Map<String, Double> sorted = budget
  9.                 .entrySet()
  10.                 .stream()
  11.                 .sorted(Map.Entry.comparingByValue(Comparator.naturalOrder()))
  12.                 .collect(Collectors.toMap(Map.Entry::getKey, Map.Entry::getValue, (e1, e2) -> e2, LinkedHashMap::new));
  13. //sorted为{期权账户=500.0, 普通账户=1000.0, 信用账户=2000.0, 理财账户=4000.0}

List< Map>根据map里某一value排序

方式一：

  1. List<Map<String, Object>> resultlist = new ArrayList<>();
  2.         Map<String, Object> map1 = new HashMap<String, Object>();
  3.         map1.put("id", "1");
  4.         map1.put("name", "张三");
  5.         map1.put("Score", 86.5);
  6.         Map<String, Object> map2 = new HashMap<String, Object>();
  7.         map2.put("id", "2");
  8.         map2.put("name", "李四");
  9.         map2.put("Score", 90.0);
  10.         Map<String, Object> map3 = new HashMap<String, Object>();
  11.         map3.put("id", "3");
  12.         map3.put("name", "王五");
  13.         map3.put("Score", 70.5);
  15.         resultlist.add(map1);
  16.         resultlist.add(map2);
  17.         resultlist.add(map3);
  19.         Collections.sort(resultlist, new Comparator<Map<String, Object>>() {
  20.             @Override
  21.             public int compare(Map<String, Object> o1, Map<String, Object> o2) {
  22.                 Double one = (Double) o1.get("Score");
  23.                 Double two = (Double) o2.get("Score");
  24.                 return one.compareTo(two);  //one.compareTo(two)为升序，two.compareTo(one)为降序
  25.             }
  26.         });
  28.         System.out.println(resultlist);
  29. //输出[{Score=70.5, name=王五, id=3}, {Score=86.5, name=张三, id=1}, {Score=90.0, name=李四, id=2}]

方式二：

  1. List<Map<String, Object>> resultlist = new ArrayList<>();
  2.         Map<String, Object> map1 = new HashMap<String, Object>();
  3.         map1.put("id", "1");
  4.         map1.put("name", "张三");
  5.         map1.put("Score", 86.5);
  6.         Map<String, Object> map2 = new HashMap<String, Object>();
  7.         map2.put("id", "2");
  8.         map2.put("name", "李四");
  9.         map2.put("Score", 90.0);
  10.         Map<String, Object> map3 = new HashMap<String, Object>();
  11.         map3.put("id", "3");
  12.         map3.put("name", "王五");
  13.         map3.put("Score", 70.5);
  15.         resultlist.add(map1);
  16.         resultlist.add(map2);
  17.         resultlist.add(map3);
  18.         Collections.sort(resultlist, (o1, o2) -> {
  19.             Double one =  Double.valueOf(o1.get("Score").toString()) ;
  20.             Double two =  Double.valueOf(o2.get("Score").toString()) ;
  21.             return one.compareTo(two);  //one.compareTo(two)为升序，two.compareTo(one)为降序
  22.         });
  23.         System.out.println(resultlist);
  24. //输出[{Score=70.5, name=王五, id=3}, {Score=86.5, name=张三, id=1}, {Score=90.0, name=李四, id=2}]

List< Object>根据Object的某一属性对List进行排序

升序：

  1. List<User> userList = new ArrayList<>(Arrays.asList(
  2.                 new User(1L,new BigDecimal("10"),18),
  3.                 new User(2L,new BigDecimal("20"),19),
  4.                 new User(3L,new BigDecimal("100"),17)));
  5.         System.out.println("排序前"+userList);
  7.         userList = userList.stream().sorted(Comparator.comparing(User::getAge)).collect(Collectors.toList());
  9.         System.out.println("排序后"+userList);

输出结果

  1. 排序前[User{id=1, money=10, age=18}, User{id=2, money=20, age=19}, User{id=3, money=100, age=17}]
  2. 排序后[User{id=3, money=100, age=17}, User{id=1, money=10, age=18}, User{id=2, money=20, age=19}]

倒序：

  1. List<User> userList = new ArrayList<>(Arrays.asList(
  2.                 new User(1L,new BigDecimal("10"),18),
  3.                 new User(2L,new BigDecimal("20"),19),
  4.                 new User(3L,new BigDecimal("100"),17)));
  5.         System.out.println("排序前"+userList);
  7.         userList = userList.stream().sorted(Comparator.comparing(User::getAge).reversed()).collect(Collectors.toList());
  9.         System.out.println("排序后"+userList);

输出结果

  1. 排序前[User{id=1, money=10, age=18}, User{id=2, money=20, age=19}, User{id=3, money=100, age=17}]
  2. 排序后[User{id=2, money=20, age=19}, User{id=1, money=10, age=18}, User{id=3, money=100, age=17}]

多条件排序:

  1. List<User> userList = new ArrayList<>(Arrays.asList(
  2.                 new User(1L,new BigDecimal("10"),18),
  3.                 new User(2L,new BigDecimal("20"),19),
  4.                 new User(3L,new BigDecimal("100"),18)));
  5.         System.out.println("排序前"+userList);
  7.         userList = userList.stream().sorted(Comparator.comparing(User::getAge).thenComparing(User::getId)).collect(Collectors.toList());
  9.         System.out.println("排序后"+userList);

输出结果

  1. 排序前[User{id=1, money=10, age=18}, User{id=2, money=20, age=19}, User{id=3, money=100, age=18}]
  2. 排序后[User{id=1, money=10, age=18}, User{id=3, money=100, age=18}, User{id=2, money=20, age=19}]

复杂List<Map<String, List<Map<String, Object>>>>排序实例

要求：不仅内部要根据Score排序，整个List也要根据Score排序

  1. List<Map<String, List<Map<String, Object>>>> tagsList = new ArrayList<>();
  3.         List<Map<String, Object>> resultlist1 = new ArrayList<>();
  4.         Map<String, List<Map<String, Object>>> listMap1 = new HashMap<>();
  5.         Map<String, Object> map1 = new HashMap<String, Object>();
  6.         map1.put("id", "1");
  7.         map1.put("name", "张一");
  8.         map1.put("Score", 7L);
  9.         Map<String, Object> map2 = new HashMap<String, Object>();
  10.         map2.put("id", "2");
  11.         map2.put("name", "李一");
  12.         map2.put("Score", 100L);
  13.         Map<String, Object> map3 = new HashMap<String, Object>();
  14.         map3.put("id", "3");
  15.         map3.put("name", "王一");
  16.         map3.put("Score", 86L);
  17.         Map<String, Object> map4 = new HashMap<String, Object>();
  18.         map4.put("id", "4");
  19.         map4.put("name", "赵一");
  20.         map4.put("Score", 0L);
  21.         resultlist1.add(map1);
  22.         resultlist1.add(map2);
  23.         resultlist1.add(map3);
  24.         resultlist1.add(map4);
  25.         listMap1.put("1", resultlist1);
  26.         tagsList.add(listMap1);
  28.         List<Map<String, Object>> resultlist2 = new ArrayList<>();
  29.         Map<String, List<Map<String, Object>>> listMap2 = new HashMap<>();
  30.         Map<String, Object> map2_1 = new HashMap<String, Object>();
  31.         map2_1.put("id", "1");
  32.         map2_1.put("name", "张二");
  33.         map2_1.put("Score", 709L);
  34.         Map<String, Object> map2_2 = new HashMap<String, Object>();
  35.         map2_2.put("id", "2");
  36.         map2_2.put("name", "李二");
  37.         map2_2.put("Score", 0L);
  38.         Map<String, Object> map2_3 = new HashMap<String, Object>();
  39.         map2_3.put("id", "3");
  40.         map2_3.put("name", "王二");
  41.         map2_3.put("Score", 456L);
  42.         Map<String, Object> map3_4 = new HashMap<String, Object>();
  43.         map3_4.put("id", "4");
  44.         map3_4.put("name", "赵二");
  45.         map3_4.put("Score", 0L);
  46.         resultlist2.add(map2_1);
  47.         resultlist2.add(map2_1);
  48.         resultlist2.add(map2_3);
  49.         resultlist2.add(map3_4);
  50.         listMap2.put("2", resultlist2);
  51.         tagsList.add(listMap2);
  53.         List<Map<String, Object>> resultlist3 = new ArrayList<>();
  54.         Map<String, List<Map<String, Object>>> listMap3 = new HashMap<>();
  55.         Map<String, Object> map3_1 = new HashMap<String, Object>();
  56.         map3_1.put("id", "1");
  57.         map3_1.put("name", "张三");
  58.         map3_1.put("Score", 9L);
  59.         Map<String, Object> map3_2 = new HashMap<String, Object>();
  60.         map3_2.put("id", "2");
  61.         map3_2.put("name", "李三");
  62.         map3_2.put("Score", 0L);
  63.         Map<String, Object> map3_3 = new HashMap<String, Object>();
  64.         map3_3.put("id", "3");
  65.         map3_3.put("name", "王三");
  66.         map3_3.put("Score", 86L);
  67.         Map<String, Object> map2_4 = new HashMap<String, Object>();
  68.         map2_4.put("id", "4");
  69.         map2_4.put("name", "赵三");
  70.         map2_4.put("Score", 920L);
  71.         resultlist3.add(map3_1);
  72.         resultlist3.add(map3_2);
  73.         resultlist3.add(map3_3);
  74.         resultlist3.add(map2_4);
  75.         listMap3.put("3", resultlist3);
  76.         tagsList.add(listMap3);
  78.         System.out.println("未排序时的tagsList：" + tagsList);
  80.         //先对每一个里的Score降序排序，排序后tagsList里的每一个元素中的List<Map<String, Object>>都是有序的
  81.         for (Map<String, List<Map<String, Object>>> ins : tagsList) {
  82.             for (List<Map<String, Object>> v : ins.values()) {
  83.                 Collections.sort(v, (o1, o2) -> {
  84.                     Long one = Long.valueOf(o1.get("Score").toString());
  85.                     Long two = Long.valueOf(o2.get("Score").toString());
  86.                     return two.compareTo(one);
  87.                 });
  88.             }
  89.         }
  90.         System.out.println("对每一个Score内部降序排序后的tagsList：" + tagsList);
  92.         Collections.sort(tagsList, (o1, o2) -> {
  93.             Long one = 0L;
  94.             Long two = 0L;
  95.             for (List<Map<String, Object>> v : o1.values()) {
  96.                 one = Long.valueOf(v.get(0).get("Score").toString());
  97.                 break;
  98.             }
  99.             for (List<Map<String, Object>> v : o2.values()) {
  100.                 two = Long.valueOf(v.get(0).get("Score").toString());
  101.                 break;
  102.             }
  103.             return two.compareTo(one);
  104.         });
  105.         System.out.println("根据最内层的最大值排序后的tagsList：" + tagsList);

输出结果

  1. 未排序时的tagsList：
  2. [{1=[{Score=7, name=张一, id=1}, {Score=100, name=李一, id=2}, {Score=86, name=王一, id=3}, {Score=0, name=赵一, id=4}]}, 
  3. {2=[{Score=709, name=张二, id=1}, {Score=709, name=张二, id=1}, {Score=456, name=王二, id=3}, {Score=0, name=赵二, id=4}]}, 
  4. {3=[{Score=9, name=张三, id=1}, {Score=0, name=李三, id=2}, {Score=86, name=王三, id=3}, {Score=920, name=赵三, id=4}]}]
  5.     
  6. 对每一个Score内部降序排序后的tagsList：
  7. [{1=[{Score=100, name=李一, id=2}, {Score=86, name=王一, id=3}, {Score=7, name=张一, id=1}, {Score=0, name=赵一, id=4}]}, 
  8. {2=[{Score=709, name=张二, id=1}, {Score=709, name=张二, id=1}, {Score=456, name=王二, id=3}, {Score=0, name=赵二, id=4}]}, 
  9. {3=[{Score=920, name=赵三, id=4}, {Score=86, name=王三, id=3}, {Score=9, name=张三, id=1}, {Score=0, name=李三, id=2}]}]
  10.     
  11. 根据最内层的最大值排序后的tagsList：
  12. [{3=[{Score=920, name=赵三, id=4}, {Score=86, name=王三, id=3}, {Score=9, name=张三, id=1}, {Score=0, name=李三, id=2}]}, 
  13. {2=[{Score=709, name=张二, id=1}, {Score=709, name=张二, id=1}, {Score=456, name=王二, id=3}, {Score=0, name=赵二, id=4}]}, 
  14. {1=[{Score=100, name=李一, id=2}, {Score=86, name=王一, id=3}, {Score=7, name=张一, id=1}, {Score=0, name=赵一, id=4}]}]

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原文链接 : https://blog.csdn.net/weixin_43296313/article/details/122844835

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