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我正在尝试编写一个名为 zip 的函数,它接受两个列表作为参数并返回一个列表,其元素是从原始列表中交替获取的
前任。(zip '(abc) '(xyz)) 应该评估为 (axbycz)
骨架解决方案:
(define (zip l1 l2) (cond ((null? l1) l2) ((null? l2) l1) (else (cons ??? (cons ??? (zip ??? ???))))))
填写???自己。:-)
???