擅长:python、mysql、java
<p>您可以使用递归来研究当每个可能的包含正确初始字符的字母都添加到运行列表中时出现的每个“分支”:</p>
<pre><code>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)
</code></pre>
<p>输出:</p>
^{pr2}$
<p>此解决方案的工作原理与宽度优先搜索类似,因为只要当前值以前没有被调用,函数<code>get_resuls</code>将继续遍历整个列表。函数看到的值被添加到<code>_seen</code>列表中,最终停止递归调用流。在</p>
<p>此解决方案还将忽略具有重复项的结果:</p>
<pre><code>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)
</code></pre>
<p>输出:</p>
<pre><code>['ant', 'tiger', 'racoon', 'ning', 'giraffe', 'elephant']
</code></pre>