)两个参数的重载方法（最简单的形式）
public static <T, K, U> Collector<T, ?, Map<K,U>> toMap(Function<? super T, ? extends K> keyMapper, Function<? super T, ? extends U> valueMapper) { return toMap(keyMapper, valueMapper, throwingMerger(), HashMap::new); }2)三个参数的重载方法（包含冲突处理）
public static <T, K, U> Collector<T, ?, Map<K,U>> toMap(Function<? super T, ? extends K> keyMapper, Function<? super T, ? extends U> valueMapper, BinaryOperator<U> mergeFunction) { return toMap(keyMapper, valueMapper, mergeFunction, HashMap::new); }3)四个参数的重载方法（指定Map实现）
public static <T, K, U, M extends Map<K, U>> Collector<T, ?, M> toMap(Function<? super T, ? extends K> keyMapper, Function<? super T, ? extends U> valueMapper, BinaryOperator<U> mergeFunction, Supplier<M> mapSupplier) { BiConsumer<M, T> accumulator = (map, element) -> map.merge(keyMapper.apply(element), valueMapper.apply(element), mergeFunction); return new CollectorImpl<>(mapSupplier, accumulator, mapMerger(mergeFunction), CH_ID); }接下来，我们结合使用示例详细讲解。
2. 使用示例
2.1 将对象的某些属性转换为Map
假设有一个城市列表，需要将其转换为Map，其中Key为城市ID、Value为城市名称，转换方法如下所示：
@Getter @Setter public class City { private Integer cityId; private String cityName; public City(Integer cityId, String cityName) { this.cityId = cityId; this.cityName = cityName; } }List<City> cityList = Arrays.asList( new City(1, "北京"), new City(2, "上海"), new City(3, "广州"), new City(4, "深圳") ); Map<Integer, String> cityMap = cityList.stream() .collect(Collectors.toMap(City::getCityId, City::getCityName)); System.out.println(cityMap);输出结果：
2.2 将对象列表转换为Map（ID -> 对象）
仍然使用上面的城市列表，需要将其转换为Map，其中Key为城市ID、Value为城市对象，转换方法如下所示：
List<City> cityList = Arrays.asList( new City(1, "北京"), new City(2, "上海"), new City(3, "广州"), new City(4, "深圳") ); Map<Integer, City> cityMap = cityList.stream() .collect(Collectors.toMap(City::getCityId, city -> city)); City city = cityMap.get(1); System.out.println("城市ID: " + city.getCityId()); System.out.println("城市名称: " + city.getCityName());输出结果如下所示：
城市ID: 1
城市名称: 北京
上面的写法等价于：
Map<Integer, City> cityMap = cityList.stream() .collect(Collectors.toMap(City::getCityId, Function.identity()));因为Function.identity()内部实现是下面这样的：
static <T> Function<T, T> identity() { return t -> t; }2.3 键冲突处理
假设上面的城市列表中有一个ID重复的城市：
List<City> cityList = Arrays.asList( new City(1, "北京"), new City(2, "上海"), new City(3, "广州"), new City(4, "深圳"), new City(4, "天津") ); Map<Integer, String> cityMap = cityList.stream() .collect(Collectors.toMap(City::getCityId, City::getCityName)); System.out.println("城市ID: 4, 城市名称: " + cityMap.get(4));此时运行代码，会抛出java.lang.IllegalStateException异常，如下图所示：
有3种常见的键冲突处理方式，分别是保留旧值、使用新值和合并值，接下来一一讲解。
1)方式一：保留旧值
Map<Integer, String> cityMap = cityList.stream() .collect(Collectors.toMap(City::getCityId, City::getCityName, (oldValue, newValue) -> oldValue));输出结果：
城市ID: 4, 城市名称: 深圳
2)方式二：使用新值
Map<Integer, String> cityMap = cityList.stream() .collect(Collectors.toMap(City::getCityId, City::getCityName, (oldValue, newValue) -> newValue));输出结果：
城市ID: 4, 城市名称: 天津
3)方式三：合并值
Map<Integer, String> cityMap = cityList.stream() .collect(Collectors.toMap(City::getCityId, City::getCityName, (oldValue, newValue) -> oldValue + ", " + newValue));输出结果：
城市ID: 4, 城市名称: 深圳, 天津
2.4 数据分组聚合
假设有一个销售记录列表，需要将其转换为Map，其中Key为销售员、Value为该销售员的总销售额，转换方法如下所示：
@Getter @Setter public class SalesRecord { private String salesPerson; private BigDecimal amount; public SalesRecord(String salesPerson, BigDecimal amount) { this.salesPerson = salesPerson; this.amount = amount; } }List<SalesRecord> salesRecordList = Arrays.asList( new SalesRecord("张三", new BigDecimal("1000")), new SalesRecord("李四", new BigDecimal("2000")), new SalesRecord("张三", new BigDecimal("980")) ); Map<String, BigDecimal> salesRecordMap = salesRecordList.stream() .collect(Collectors.toMap(SalesRecord::getSalesPerson, SalesRecord::getAmount, BigDecimal::add)); System.out.println(salesRecordMap);输出结果：
上面的例子是销售额累加，也可以只取最小值：
Map<String, BigDecimal> salesRecordMap = salesRecordList.stream() .collect(Collectors.toMap(SalesRecord::getSalesPerson, SalesRecord::getAmount, BigDecimal::min));此时的输出结果：
或者只取最大值：
Map<String, BigDecimal> salesRecordMap = salesRecordList.stream() .collect(Collectors.toMap(SalesRecord::getSalesPerson, SalesRecord::getAmount, BigDecimal::max));此时的输出结果：
2.5 指定Map实现
默认情况下，Collectors.toMap是将结果收集到HashMap中，如果有需要，我们也可以指定成TreeMap或者LinkedHashMap。
如果想要保持插入顺序，可以指定使用LinkedHashMap：
List<City> cityList = Arrays.asList( new City(2, "上海"), new City(1, "北京"), new City(4, "深圳"), new City(3, "广州") ); Map<Integer, String> cityMap = cityList.stream() .collect(Collectors.toMap(City::getCityId, City::getCityName, (existing, replacement) -> existing, LinkedHashMap::new)); System.out.println(cityMap);输出结果：
如果想要按键排序，可以指定使用TreeMap：
List<City> cityList = Arrays.asList( new City(2, "上海"), new City(1, "北京"), new City(4, "深圳"), new City(3, "广州") ); Map<Integer, String> cityMap = cityList.stream() .collect(Collectors.toMap(City::getCityId, City::getCityName, (existing, replacement) -> existing, TreeMap::new)); System.out.println(cityMap);输出结果：
)
