itertools mode 之 combinations用法
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itertools mode 之 combinations用法
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itertools mode 中 combinations的功能:
from itertools import combinations
print list(combinations(list,r))
將打印 list中的所有長度為r的,子集的合集
e.g.
from itertools import combinations
list = [1,2,3]
print (list(combinations(list,2)))
>>>
[(1,2),(1,3),(2,3)]
所以當遇到排列組合的問題時,combinations會很好用
思路:利用combinations列出符合長度k的子集,再篩選出合為n的子集
from itertools import combinations
class Solution(object):
??? def combinationSum3(self, k, n):
return [res for res in combinations(range(1,10),k) if sum(res) == n]
詳細點的寫法:
from itertools import combinations
class Solution(object):
??? def combinationSum3(self, k, n):
res = []
raw = combinations(range(1,10),k)
for i in raw:
if sum(i) == n:
res.append(i)
return res
from itertools import combinations
print list(combinations(list,r))
將打印 list中的所有長度為r的,子集的合集
e.g.
from itertools import combinations
list = [1,2,3]
print (list(combinations(list,2)))
>>>
[(1,2),(1,3),(2,3)]
所以當遇到排列組合的問題時,combinations會很好用
leetcode例題:
216. Combination Sum III
Find all possible combinations of k numbers that add up to a number n, given that only numbers from 1 to 9 can be used and each combination should be a unique set of numbers.
Note:
- All numbers will be positive integers.
- The solution set must not contain duplicate combinations.
Example 1:
Input: k = 3, n = 7 Output: [[1,2,4]]Example 2:
Input: k = 3, n = 9 Output: [[1,2,6], [1,3,5], [2,3,4]]思路:利用combinations列出符合長度k的子集,再篩選出合為n的子集
from itertools import combinations
class Solution(object):
??? def combinationSum3(self, k, n):
return [res for res in combinations(range(1,10),k) if sum(res) == n]
詳細點的寫法:
from itertools import combinations
class Solution(object):
??? def combinationSum3(self, k, n):
res = []
raw = combinations(range(1,10),k)
for i in raw:
if sum(i) == n:
res.append(i)
return res
轉載于:https://www.cnblogs.com/phinza/p/10231257.html
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