[Leetcode]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] 

Combination Sum 的扩展题~这题中单个元素不能重复使用~解法也是差不多的，注意递归的时候传进去的start应该是当前元素的下一个,还有注意判断重复元素时用的是i > start，Combination Sum里用的是i > 0，因为虽然一个元素不可以重复使用，但是如果这个元素重复出现是允许的，但是为了避免出现重复的结果集，我们只对于第一次得到这个数进行递归，接下来就跳过这个元素了，因为接下来的情况会在上一层的递归函数被考虑到，这样就可以避免重复元素的出现（参考自

http://blog.csdn.net/linhuanmars/article/details/20829099）

class Solution:    # @param candidates, a list of integers    # @param target, integer    # @return a list of lists of integers    def combinationSum2(self, candidates, target):        if candidates is None or len(candidates) == 0: return []        self.res = []        candidates.sort()        self.helper(candidates, target, 0, [])        return self.res            def helper(self, candidates, remainder, start, item):        if remainder == 0:            self.res.append(item[:])            return        if remainder < 0: return        for i in xrange(start, len(candidates)):            if i > start and candidates[i] == candidates[i - 1]:                continue            item.append(candidates[i])            self.helper(candidates, remainder - candidates[i], i + 1, item)            item.pop()