(编辑：jimmy 日期: 2025/7/3 浏览：2)
使用递归实现
words = ['giraffe', 'elephant', 'ant', 'tiger', 'racoon', 'cat', 'hedgehog', 'mouse']
def get_results(_start, _current, _seen):
if all(c in _seen for c in words if c[0] == _start[-1]):
yield _current
else:
for i in words:
if i[0] == _start[-1]:
yield from get_results(i, _current+[i], _seen+[i])
new_d = [list(get_results(i, [i], []))[0] for i in words]
final_d = max([i for i in new_d if len(i) == len(set(i))], key=len)
输出：
['hedgehog', 'giraffe', 'elephant', 'tiger', 'racoon']
工作原理类似于广度优先搜索，因为只要当前值之前没有被调用，get_results函数就会继续遍历整个列表。函数已经查找过的值被添加到_seen列表中，最终停止递归调用流。这个解决方案也会忽略重复的结果，
words = ['giraffe', 'elephant', 'ant', 'ning', 'tiger', 'racoon', 'cat', 'hedgehog', 'mouse',] new_d = [list(get_results(i, [i], []))[0] for i in words] final_d = max([i for i in new_d if len(i) == len(set(i))], key=len)
输出：
['ant', 'tiger', 'racoon', 'ning', 'giraffe', 'elephant']