Combination Sum
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:
For example, given candidate set 2,3,6,7
and target 7
,
A solution set is:
[7]
[2, 2, 3]
分析:
又是一题深度搜索题(回溯题)。难度在剪枝。
刚刚开始我的搜索函数是这样的:
代码:
def dfs(self, candidates, target, knums, solutions): sum_knums = sum(knums) if sum_knums == target: return knums for i in candidates: all_sum = sum_knums + i if all_sum <= target: res = self.dfs(candidates, target, knums + [i], solutions) if res: solutions.append(res)
knums表示当前程序中的解,保存添加进去的元素。
solutions表示最终程序所有合法的解。
搜索中,先判断当前解是否大于target,如果大于则终止判断,该解无效,如果等于target,则为有效解,如果小于target,那么依次往knums中添加元素,再次搜索。
这个函数效率很差,每一次搜索时都会先从candidates数组的第一个元素开始尝试,得到的solutions中有重复的解。比如题目中给的例子,得到的solutions的值是这样子的:[[2, 2, 3], [2, 3, 2], [3, 2, 2], [7]],此时solutions还需要去重才能得到我们最后的答案。
我们发现,如果在向当前解中添加一个元素时,如果能够只添加比当前元素大或者相等的元素,而比当前元素小得元素则不再尝试的话,即可消除重复的解。
修改代码得:
def dfs(self, candidates, target, knums, cur, solutions): sum_knums = sum(knums) if sum_knums == target: solutions.append(knums) curcandidates = [i for i in candidates if i >= cur] for i in curcandidates: all_sum = sum_knums + i if all_sum <= target: self.dfs(candidates, target, knums + [i], i, solutions)
我们发现,如果每次递归时,不是给定当前的元素值,而是返回当前元素所在的下标时即可取消掉列表生成表达式。
代码:
def dfs(self, candidates, target, start, end, knums, solutions): sum_knums = sum(knums) if sum_knums == target: solutions.append(knums) i = start while i <= end: all_sum = sum_knums + candidates[i] if all_sum <= target: self.dfs(candidates, target, i, end, knums + [candidates[i]], solutions) i += 1
取消掉了列表生成,效率进一步提高,但是,每次递归都会执行sum函数,这个函数的执行需要时间,有没有办法能够取消掉sum函数的执行呢?
上面的代码,我们判断停止的条件是让当前解的总和等于target,如果我们的判断条件修改一下,那么可以做到取消掉sum函数。
代码:
class Solution(object): def combinationSum(self, candidates, target): """ :type candidates: List[int] :type target: int :rtype: List[List[int]] """ solutions = [] candidates.sort() self.dfs(candidates, target, 0, len(candidates) - 1, list(), solutions) return solutions def dfs(self, candidates, gap, start, end, knums, solutions): # 在本搜索中去掉了累加 if gap == 0: solutions.append(knums) return i = start while i <= end: if gap < candidates[i]: return self.dfs(candidates, gap - candidates[i], i, end, knums + [candidates[i]], solutions) i += 1