为什么在Python递归中我的列表中得到了None？

huangapple
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英文:

why am i getting None in my list in Recursion in python

问题

我的 Slack 输出为 slack=[[],[],[]]，为什么输出没有追加进去呢？

  1. class Solution:
  2.     def __init__(self):
  3.         self.slack = []
  5.     def combinationSum2(self, candidates: List[int], target: int) -> List[List[int]]:
  6.         output = []
  7.         i = 0
  9.         def rec(i, target, output, candidates):
  10.             if i >= len(candidates):
  11.                 if target == 0:
  12.                     self.slack.append(output[:])  # 需要使用副本，而不是直接追加
  13.                 return self.slack
  15.             if candidates[i] <= target:
  16.                 output.append(candidates[i])
  17.                 rec(i + 1, target - candidates[i], output, candidates)
  18.                 output.pop()
  19.             rec(i + 1, target, output, candidates)
  21.         rec(i, target, output, candidates)
  22.         return self.slack

需要在self.slack.append(output[:])中使用副本来追加输出，而不是直接追加 output，以确保正确追加结果。

英文:

my slack is getting o/p as slack=[[],[],[]] output is not appending in it why?
candidates = [10,1,2,7,6,1,5], target = 8 Output: [ [1,1,6], [1,2,5], [1,7], [2,6] ]

  1. type here
  2. class Solution:
  3.     def __init__(self):
  4.         self.slack =[]
  6.     def combinationSum2(self, candidates: List[int], target: int) -&gt; List[List[int]]:
  7.         output =[]
  8.         i =0
  9.         def rec(i,target,output,candidates):
  10.             if i&gt;=len(candidates):
  11.                 if target == 0: 
  12.                     self.slack.append(output) 
  13.                 return self.slack
  14.             
  15.             if candidates[i]&lt;=target:
  16.                 output.append(candidates[i])
  17.                 rec(i+1,target-candidates[i],output,candidates)
  18.                 output.pop()
  19.             rec(i+1,target,output,candidates)
  20.         
  21.         rec(i,target,output,candidates)
  22.         return self.slack

output needs to be appended inside the slack

答案1

得分: 1

  1. from typing import List
  3. class Solution:
  4.     def __init__(self):
  5.         self.slack = []
  7.     def combinationSum2(self, candidates: List[int], target: int) -> List[List[int]]:
  8.         output = []
  9.         i = 0
  10.         candidates.sort()  # 对候选项进行排序以处理重复项
  12.         def rec(i, target, output, candidates):
  13.             if i >= len(candidates):
  14.                 if target == 0:
  15.                     self.slack.append(output.copy())  # 将 'output' 的副本附加到 'self.slack'
  16.                 return self.slack
  18.             if candidates[i] <= target:
  19.                 output.append(candidates[i])
  20.                 rec(i + 1, target - candidates[i], output, candidates)
  22.                 # 跳过重复项
  23.                 while i + 1 < len(candidates) and candidates[i] == candidates[i + 1]:
  24.                     i += 1
  26.                 output.pop()  # 回溯时移除最后一个元素
  27.             rec(i + 1, target, output, candidates)
  29.         rec(i, target, output, candidates)
  30.         return self.slack
  32. # 测试
  33. candidates = [10, 1, 2, 7, 6, 1, 5]
  34. target = 8
  36. solution = Solution()
  37. result = solution.combinationSum2(candidates, target)
  38. print(result)  # [[1, 1, 6], [1, 2, 5], [1, 7], [2, 6]]
  39. assert result == [[1, 1, 6], [1, 2, 5], [1, 7], [2, 6]]
英文:

Try this

  1. from typing import List
  3. class Solution:
  4.     def __init__(self):
  5.         self.slack = []
  7.     def combinationSum2(self, candidates: List[int], target: int) -&gt; List[List[int]]:
  8.         output = []
  9.         i = 0
  10.         candidates.sort()  # Sort the candidates list to handle duplicates
  12.         def rec(i, target, output, candidates):
  13.             if i &gt;= len(candidates):
  14.                 if target == 0:
  15.                     self.slack.append(output.copy())  # Append a copy of &#39;output&#39; to &#39;self.slack&#39;
  16.                 return self.slack
  18.             if candidates[i] &lt;= target:
  19.                 output.append(candidates[i])
  20.                 rec(i + 1, target - candidates[i], output, candidates)
  22.                 # Skip duplicates
  23.                 while i + 1 &lt; len(candidates) and candidates[i] == candidates[i + 1]:
  24.                     i += 1
  26.                 output.pop()  # Remove the last element as we backtrack
  27.             rec(i + 1, target, output, candidates)
  29.         rec(i, target, output, candidates)
  30.         return self.slack

Testing

  1. candidates = [10, 1, 2, 7, 6, 1, 5]
  2. target = 8
  4. solution = Solution()
  5. result = solution.combinationSum2(candidates, target)
  6. print(result) # [[1, 1, 6], [1, 2, 5], [1, 7], [2, 6]]
  7. assert result ==  [[1,1,6], [1,2,5], [1,7], [2,6]]

huangapple
  • 本文由 发表于 2023年7月23日 16:32:11
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  • python
  • recursion
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