链接:
题目:
给定一个无重复元素的数组 candidates 和一个目标数 target ,找出 candidates 中所有可以使数字和为 target 的组合。candidates 中的数字可以无限制重复被选取。
说明:
所有数字(包括 target)都是正整数。
解集不能包含重复的组合。
示例 1:
输入: candidates = [2,3,6,7], target = 7,
所求解集为:
[
[7],
[2,2,3]
]
示例 2:
输入: candidates = [2,3,5], target = 8,
所求解集为:
[
[2,2,2,2],
[2,3,3],
[3,5]
]
别人的解法:回溯
class Solution(object):
def combinationSum(self, candidates, target):
res = []
candidates.sort()
self.dfs(candidates, target, 0, [], res)
return res
def dfs(self, nums, target, index, path, res):
if target < 0:
return # backtracking
if target == 0:
res.append(path)
return
for i in xrange(index, len(nums)):
self.dfs(nums, target-nums[i], i, path+[nums[i]], res)
我的改进解法:44ms,超越99.8%。把判断放在循环里,更少调用函数
class Solution(object):
def combinationSum(self, candidates, target):
candidates.sort()
result = []
self.combine_sum(candidates, 0, [], result, target)
return result
def combine_sum(self, nums, start, path, result, target):
if target == 0:
result.append(path)
return
for i in range(start, len(nums)):
if nums[i] > target:
break
return
self.combine_sum(nums, i, path+[nums[i]], result, target-nums[i])