[LeetCode]题解（python）：040-Combination Sum II

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题目来源

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https://leetcode.com/problems/combination-sum-ii/

Given a collection of candidate numbers (C) and a target number (T), find all unique combinations in C where the candidate numbers sums to T.

Each number in C may only be used once in the combination.

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题意分析

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Input: a list as candidates, a value named target

Output:the list number that sumed to target

Conditions:在list里面找若干个数，使得和为target，注意每个数可以取一次

Note:

• All numbers (including target) will be positive integers.
• Elements in a combination (a1, a2, … , ak) must be in non-descending order. (ie, a1 ≤ a2 ≤ … ≤ ak).
• The solution set must not contain duplicate combinations.

For example, given candidate set 10,1,2,7,6,1,5 and target 8, 
A solution set is: 
[1, 7] 
[1, 2, 5] 
[2, 6] 
[1, 1, 6]

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题目思路

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与上题类似，先对list进行排序，然后穷举即可，注意可能会出现重复情况，所以需要去重（加一个判断语句），另外注意传递时的参数的范围

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AC代码（Python）

 1 _author_ = "YE"
 2 # -*- coding:utf-8 -*-
 3
 4 class Solution(object):
 5     def find(self,candidates, target, start, valueList):
 6         if target == 0:
 7             if valueList not in Solution.ans:
 8                 Solution.ans.append(valueList)
 9         length = len(candidates)
10         for i in range(start, length):
11             if candidates[i] > target:
12                 return
13             self.find(candidates, target - candidates[i], i + 1, valueList + [candidates[i]])
14
15     def combinationSum2(self, candidates, target):
16         """
17         :type candidates: List[int]
18         :type target: int
19         :rtype: List[List[int]]
20         """
21         candidates.sort()
22         Solution.ans = []
23         self.find(candidates, target, 0, [])
24         return Solution.ans
25
26 candidates = [10,1,2,7,6,1,5]
27 target = 8
28 s = Solution()
29 print(s.combinationSum2(candidates,target))