Java 函数式编程
Java 8 Streams API 详解
写链式调用
参考博客：
https://cloud.tencent.com/developer/article/1591343
https://www.cnblogs.com/dengchengchao/p/11855723.html
//求只出现一次的数字(其余数字出现都是2次)
//比如，求1,2,2,3,3中出现一次的数字
import java.util.*;
import java.util.Arrays;
import java.util.List;
import java.util.stream.Collectors;
public class Solution {
/**
* @param nums: Represents the incoming number
* @return: A int value representing whether the
* return is a palindrome or not
*/
public int uniqueNumber(int nums[]) {
// write your code here
List<Integer> numlist = Arrays.stream(nums).boxed().collect(Collectors.toList());
int result = numlist.stream().reduce((a,b) -> a ^ b ).get();
return result;
}
}
//得到字符串数组的各个字符串长度
import java.util.*;
import java.util.stream.*;
public class Main {
public static void main(String[] args) {
List<String> strings = Arrays.asList(
"abc", "", "bc", "12345",
"efg", "abcd","", "jkl");
List<Integer> lengths = strings
.stream()
.filter(string -> !string.isEmpty())
.map(s -> s.length())
.collect(Collectors.toList());
lengths.forEach((s) -> System.out.println(s));
}
}
//求出1~100中所有偶数的和
import java.util.*;
import java.util.stream.IntStream;
public class Main {
public static void main(String[] args) {
// write your code here
int result=IntStream.range(1,100+1)
.filter(x-> x%2==0)
.reduce(0,(a,b)->a+b);
System.out.println("Even sum of 1 - 100: "+result);
}
}
//韩信点兵，三人一组余两人，五人一组余三人，七人一组余四人，请问最少需要多少士兵。
//无脑解法，直接暴力
import java.util.*;
import java.util.stream.IntStream;
public class Main {
public static void main(String[] args) {
// write your code here
IntStream.range(1,1000000000)
.filter(x-> (x%3==2&&x%5==3&&x%7==4))
.findFirst()
.ifPresent(e -> System.out.println(e));
}
}
//"[1, 2, 3]"这样的字符串转 int[]
import java.util.*;
public class Solution {
public int[] arrayConversion(String str) {
// -- write your code here --
int [] a = Arrays.stream(str.substring(1,str.length()-1).split(", "))
.mapToInt(Integer::parseInt)
.toArray();
return a;
}
}
//数组去重后求和
//下面这种写法贼慢
import java.util.*;
import java.util.List;
import java.util.stream.Collectors;
public class Solution {
public int getSum(int arr[]) {
List<Integer> list1 = Arrays.stream(arr).boxed().collect(Collectors.toList());
List<Integer> newList = list1.stream()
.distinct()
.collect(Collectors.toList());
return newList.stream()
.reduce(0,(a,b)->a+b);
}
}
//java8 stream求平均值，最大值，最小值，求和
import java.util.Arrays;
import java.util.List;
import java.util.stream.Collectors;
public class Calculate {
public static void main(String[] args) {
/*
* 不允许list中存在为空的值，不然会异常
*/
List<Double> list = Arrays.asList(0D, 1D, 1.2, 5.6);
List<Integer> integers = Arrays.asList(0, 1, 2, 56);
Double avg = list.stream().collect(Collectors.averagingDouble(Double::doubleValue));
Double avgs = list.stream().mapToDouble(Double::doubleValue).average().orElse(0D);
Double intAvg = integers.stream().mapToInt(Integer::intValue).average().orElse(0D);
Double min = list.stream().min(Double::compareTo).get();
Double max = list.stream().max(Double::compareTo).get();
Double sum = list.stream().mapToDouble(Double::doubleValue).sum();
Double sums = list.stream().reduce(Double::sum).get();
}
}
借助Java8实现柯里化
Java中的柯里化看起来是没有语法糖的，至少我没有找到。。。
参考博客：
https://www.jianshu.com/p/c623b8b2aec8 里面提到了4种方式实现柯里化
import java.util.*;
import java.util.function.*;
public class Main {
public static void main(String[] args) {
IntFunction<IntFunction<IntUnaryOperator>> f = x -> y -> z -> (x + y) * z;
IntFunction<IntUnaryOperator> g = f.apply(4);
IntUnaryOperator mulBy9 = g.apply(5);
System.out.println(mulBy9.applyAsInt(6)); //54
}
}
使用Vavr
参考：
https://zhuanlan.zhihu.com/p/194317390 VAVR：颠覆你的 Java 体验-知乎
https://www.jianshu.com/p/f0106fb32404 Java8+ 函数库Vavr简介-简书
