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

The same repeated number may be chosen from C unlimited number of times.

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 2,3,6,7 and target 7, 
A solution set is: 
[7] 
[2, 2, 3]

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。所有的数字可以使用无限次，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：

　　思路是循环拿每个数字去迭代，找出所有的组合，满足条件返回，不满足退回上一步继续下一个数字的尝试。看特征，仍然是用回溯法的思路去做。

AC代码：

class Solution(object):

    def combinationSum(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)):

                    backtracing(target - candidates[i], candidates, i, temp_list + [candidates[i]])

        backtracing(target, sorted(list(set(candidates))), 0, [])

        return ret_list