【LeetCode题意分析&解答】40. 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.

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]

题意分析：

　　本题是给定一个数字集合（可能有重复）和一个目标值，让你找出所有可能的组合，使之和为目标值。所有的数字只能使用一次，要求得到的组合不能有重复，并且组合里面的数字必须是升序。

解答：

　　对比一下【LeetCode题意分析&解答】39. Combination Sum，唯一的区别在于本题数字只能使用一次，而不是无限次。所以我们只需要对上一题的代码稍作修改，就可以实现本题的要求。

AC代码：

class Solution(object):
def combinationSum2(self, candidates, target):
ret_list = []
def backtracing(target, candidates, index, temp_list):
if target == 0:
ret_list.append(temp_list)
elif target > 0:
for i in xrange(index, len(candidates)):
# remove duplicates
if i != index and candidates[i] == candidates[i - 1]:
continue
backtracing(target - candidates[i], candidates, i + 1, temp_list + [candidates[i]])

backtracing(target, sorted(candidates), 0, [])
return ret_list