本文共 1946 字，大约阅读时间需要 6 分钟。
list在Groovy中很重要，记录一下list的一些见操作:
def list = []list << 2// list >> 2 // invalidassert list.size() == 1assert list == [2]//传统添加list.add(3)assert list.size() == 2 //groovy中统一的size 方法assert list == [2,3]list << "a"assert list.size() == 3assert list == [2,3,"a"] //groovy中不限制list中的数据类型，因为背后都是Objectdef other_list = [3]assert list - other_list == [2,"a"] // list可以直接相减def other_list_a = [3,"b"]assert list - other_list_a == [2,"a"] // list可以直接相减assert list * 2 == [2,3,"a",2,3,"a"] // list可以乘以一个数，达到复制效果assert list.intersect(other_list) == [3] // 取交集assert list.reverse() == ["a",3,2]assert !list.disjoint(other_list_a) // 判断有无交集list.each {item -> println(item)}assert list.collect({it*2}) == [4,6,"aa"] // mapdef list_items = [1,2,3]assert list_items.inject {x,y -> x * y} == 6 // reduce or foldassert list_items.findAll {it % 2 == 0} == [2] //取值为偶数元素assert list_items.join(",") == "1,2,3" // join 有时候很有用def complex_list = [1,2,[3,4]]assert complex_list.flatten() == [1,2,3,4]//indexingdef long_list = [1,2,3,4,5,6,7,8]assert long_list[1..3] == [2,3,4]assert long_list[-1] == 8assert long_list[-1..-3] == [8,7,6]assert long_list[-1..0] == long_list.reverse() // 反转assert long_list[-1..-8] == long_list.reverse() // 反转assert long_list[1,2,4] == [2,3,5]assert long_list[0..<2] == [1,2]
下面尝试一个简单的快排算法：
/** * @author ：ZackZhou * @date ：Created in 2020/9/1 5:42 PM * @description ：quick sort * @modified By： * @version: */def quickSort(List input_list){ if(input_list.size() < 2) return input_list int middle = input_list.size().intdiv(2) def middle_element = input_list.get(middle) def middle_list = input_list.findAll{item -> item == middle_element} def left_list = input_list.findAll{item -> item < middle_element} def right_list = input_list.findAll{item -> item > middle_element} return (quickSort(left_list) + middle_list + quickSort(right_list))}def raw_list = [3,24,5,6,6,7,82,6,7,2]quickSort(raw_list).each {println(it)}
转载地址：http://papaz.baihongyu.com/