<p>以下是一个函数中的两种解决方案,无需任何外部库:</p>
<pre><code>A = [0,2,5,6]
B = [5,6,8,9]
C= [6,7,8,9]
def finding_sequence(list_1):
sub_list = []
for j, i in enumerate(list_1):
try:
if list_1[j] - list_1[j - 1] == 1:
sub_list.append((list_1[j - 1], list_1[j]))
else:
sub_list.append('_pos')
except IndexError:
pass
sub_final_result = []
check_result=[]
if '_pos' not in sub_list[1:]:
for i in sub_list[1:]:
for k in i:
if k not in sub_final_result:
sub_final_result.append(k)
check_result.append(k)
else:
for i in sub_list:
if i != '_pos':
sub_final_result.append(i)
for i1 in i:
check_result.append(i1)
for i1 in list_1:
if i1 not in check_result:
sub_final_result.append([i1])
return sub_final_result
</code></pre>
<p>测试用例:</p>
<pre><code>print(finding_sequence(A))
</code></pre>
<p>输出:</p>
<pre><code>[(5, 6), [0], [2]]
</code></pre>
<p>第二</p>
<pre><code>print(finding_sequence(B))
</code></pre>
<p>输出:</p>
<pre><code>[(5, 6), (8, 9)]
</code></pre>
<p>附言:一个要求是:如果我的答案对你有帮助,不要接受,就用它吧。你知道吗</p>