文章作者:Tyan 博客:noahsnail.com | CSDN | 简书
1. Description
2. Solution
**解析:**Version 1,采用广度优先搜索,以列表中的每个元素为起点,寻找其后大于等于当前元素的数值,添加到当前列表中,并将当前列表添加到结果中,以当前列表以及下一个元素为起点,进行下一次广度优先搜索。最后,由于存在重复元素,因此需要对嵌套列表进行去重。
- Version 1
class Solution:
def findSubsequences(self, nums: List[int]) -> List[List[int]]:
result = []
for i in range(len(nums)):
self.bfs(nums, i 1, [nums[i]], result)
return [list(t) for t in set(tuple(_) for _ in result)]
def bfs(self, nums, start, pre, result):
for i in range(start, len(nums)):
if nums[i] >= pre[-1]:
temp = pre [nums[i]]
result.append(temp)
self.bfs(nums, i 1, temp, result)
Reference
- https://leetcode.com/problems/increasing-subsequences/