输入
ls = [0, 1, 3, 4, 7]
期望输出
[[0, 1], [3, 4], [7]]

方法一

back = 0  # 每段开始位置
result = []
for i in range(1, len(ls)+1):# 超出范围报错try:# 不连续的if ls[i-1] + 1 != ls[i]:result.append(ls[back: i])back = i  # 更新每段开始位置except:result.append(ls[back:])
print(result)

方法二

from itertools import groupby
result = []
for _, i in groupby(enumerate(ls), lambda x: x[1]-x[0]):result.append([j for _, j in i])
print(result)