java 8 stream 提供了下面幾種類型的求和
Stream::mapToInt
Stream::mapToDouble
Stream::mapToLong
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public void test() {List<Person> people = new ArrayList<>();people.add(new Person("zhangsan",20));people.add(new Person("lisi", 26));people.add(new Person("wangwu",35));int sum = people.stream() .mapToInt(p -> p.getAge()) .sum();System.out.println("Total of ages " + sum);} |
但是沒有BigDecimal類型，可以使用下面方法實現
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public void test() {List<Person> list = new ArrayList<>();list.add(new Person("zhangsan", 20, new BigDecimal(10.5)));list.add(new Person("lisi", 26, new BigDecimal(22.2)));list.add(new Person("wangwu", 35, new BigDecimal(15.54)));BigDecimal amounts = list.stream().map(item -> item.getAmount()) .reduce(BigDecimal.ZERO, BigDecimal::add);amounts = amounts.setScale(1, BigDecimal.ROUND_DOWN);System.out.println("Total of amounts:" + amounts);// 或者BigDecimal sum = list .stream() .map(Person::getAmount) .reduce(BigDecimal::add) .get(); sum = sum.setScale(1, BigDecimal.ROUND_DOWN);System.out.println("Total of sums: " + sum);} |
補充知識：Java算法——求1到100累加的和，3種循環
1、for循環
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int sum = 0;for (int i = 1; i <= 100; i++) {sum += i;}System.out.println("1到100累加的和為：" + sum); |
2、while循環
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int sum = 0;int i = 1;while (i <= 100) {sum += i;i++;}System.out.println("1到100累加的和為：" + sum); |
3、do...while循環
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int sum = 0;int i = 1;do {sum += i;i++;} while (i <= 100);System.out.println("1到100累加的和為：" + sum); |
以上這篇Java lambda 循環累加求和代碼就是小編分享給大家的全部內容了，希望能給大家一個參考，也希望大家多多支持服務器之家。
原文鏈接：https://blog.csdn.net/angus_Lucky/article/details/85062728
