39 leetcode - Combination Sum II

#!/usr/bin/python# -*- coding: utf-8 -*-'''Combination Sum IIGiven 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.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]]不能重复使用一个元素,结果不能重复'''class Solution(object):    def combinationSum2(self, candidates, target):        """        :type candidates: List[int]        :type target: int        :rtype: List[List[int]]        """        length = len(candidates)        if length == 0:            return []        candidates.sort()  #先排序        ret = []        tmp = range(length)#因为元素不能重复使用,最长为candidates的长度        self.__combinationSum2(candidates,0,length,0,target,ret,tmp,0)        return ret    def __combinationSum2(self,candidates,start,length,sum,target,ret,tmp,tmp_cur):        if sum == target:            ret.append(tmp[:tmp_cur])            return        if start == length:            return        while start < length:            if sum + candidates[start] > target:                break            tmp[tmp_cur] = candidates[start]            self.__combinationSum2(candidates,start + 1,length,sum + candidates[start],target,ret,tmp,tmp_cur + 1)            while start + 1 < length and candidates[start + 1] == candidates[start]:#结果去重,因为前面已经将重复的元素都组合过了,再组合只会导致重复结果,所以重复的元素不再组合                start += 1            start += 1if __name__ == "__main__":    s = Solution()    print s.combinationSum2([10,1,2,7,6,1,5],8)    print s.combinationSum2([1],1)