LeetCode#40 Combination Sum II

Problem Definition：

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, … , a_k) must be in non-descending order. (ie, a₁ ≤ a₂ ≤ … ≤ a_k).
• 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]

Solution：这题跟上一题的区别在于，待选的元素可能重复出现，但是一个元素不能被多次使用。所以，如果C中有两个1，那在一个解中，最多出现两个1，它们是不同的1。

所以这里要处理重复解的问题，比如上个例子中，实际上有两个潜在的[1,7]，但是最终只能用一个。

可以用HashMap来处理重复（预防多次加入同一种解）saves a lot of time。

    # @param {integer[]} candidates
    # @param {integer} target
    # @return {integer[][]}
    def combinationSum2(self, candidates, target):
        candidates.sort()
        res={}
        self.cur(candidates, target, 0, [], res)
        return res.values()
        
    def cur(self, nums, target, index, localArr, res):
        if target==0:
            key=str(localArr)
            if key not in res:
                res[key]=localArr[:]
        else:
            for i in range(index, len(nums)):
                nt=target-nums[i]
                if nt>=0:
                    localArr.append(nums[i])
                    self.cur(nums, nt, i+1, localArr, res)
                    localArr.pop()
                else:
                    break