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"""
File: subset_sum_i_naive.py
Created Time: 2023-06-17
Author: krahets (krahets@163.com)
"""
def backtrack (
state : list [ int ] ,
target : int ,
total : int ,
choices : list [ int ] ,
res : list [ list [ int ] ] ,
) :
""" 回溯演算法：子集和 I """
# 子集和等於 target 時，記錄解
if total == target :
res . append ( list ( state ) )
return
# 走訪所有選擇
for i in range ( len ( choices ) ) :
# 剪枝：若子集和超過 target ，則跳過該選擇
if total + choices [ i ] > target :
continue
# 嘗試：做出選擇，更新元素和 total
state . append ( choices [ i ] )
# 進行下一輪選擇
backtrack ( state , target , total + choices [ i ] , choices , res )
# 回退：撤銷選擇，恢復到之前的狀態
state . pop ( )
def subset_sum_i_naive ( nums : list [ int ] , target : int ) - > list [ list [ int ] ] :
""" 求解子集和 I（ ） """
state = [ ] # 狀態（子集）
total = 0 # 子集和
res = [ ] # 結果串列（子集串列）
backtrack ( state , target , total , nums , res )
return res
""" Driver Code """
if __name__ == " __main__ " :
nums = [ 3 , 4 , 5 ]
target = 9
res = subset_sum_i_naive ( nums , target )
print ( f " 輸入陣列 nums = { nums } , target = { target } " )
print ( f " 所有和等於 { target } 的子集 res = { res } " )
print ( f " 請注意，該方法輸出的結果包含重複集合 " )
