leetcode -- Combination Sum II -- 重点

https://leetcode.com/problems/combination-sum-ii/

与Combination Sum一样，只是每个元素只能用一次。子节点集合不包括本身就行。

还是要排序，排序的原因要知道：1，方便剪枝，2. 题目要求 combination要从小到大

class Solution(object):    def dfs(self, candidates, start, end, target, subres, res):        #print subres        if sum(subres) == target:            if subres not in res:                res.append(subres[:])                return        else:            i = start            while i < end:                if sum(subres) + candidates[i] <= target:                    self.dfs(candidates, i + 1, end, target, subres + [candidates[i]], res)                else:# 因为排序过，只要碰到有子节点加上之后大于target，那么就不需要继续搜后续的子节点了。因为都比这个子节点大                    return                 i += 1    def combinationSum2(self, candidates, target):        """        :type candidates: List[int]        :type target: int        :rtype: List[List[int]]        """        res = []        self.dfs(sorted(candidates), 0, len(candidates), target, [], res)        return res

自己重写code

class Solution(object):    def dfs(self, candidates, start, end, target, subres, res):        cur_sum = sum(subres)        if cur_sum == target and subres not in res:            res.append(subres)        for i in xrange(start, end):            if cur_sum + candidates[i] <= target:#这里要有等于号                self.dfs(candidates, i+1, end, target, subres + [candidates[i]], res)    def combinationSum2(self, candidates, target):        """        :type candidates: List[int]        :type target: int        :rtype: List[List[int]]        """        candidates.sort()        res = []        self.dfs(candidates, 0, len(candidates), target, [], res)        return res