040. 组合总和 II
给定一个数组 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 中的每个数字在每个组合中只能使用一次。
注意:解集不能包含重复的组合。
示例 1:
输入: candidates = [10,1,2,7,6,1,5], target = 8,
输出:
[
[1,1,6],
[1,2,5],
[1,7],
[2,6]
]
示例 2:
输入: candidates = [2,5,2,1,2], target = 5,
输出:
[
[1,2,2],
[5]
]
提示:
1 <= candidates.length <= 100
1 <= candidates[i] <= 50
1 <= target <= 30
来源:力扣(LeetCode)
链接:https://leetcode-cn.com/problems/combination-sum-ii
著作权归领扣网络所有。商业转载请联系官方授权,非商业转载请注明出处。
代码:
from leetcode_python.utils import *
class Solution:
def __init__(self):
pass
def dfs(self,listnow,sumnow,idnow):
if sumnow==self.target:
self.res.append(listnow.copy())
elif sumnow>self.target or idnow==self.length:
return
else:
for next in range(idnow,self.length):
if next>idnow and self.candidates[next]==self.candidates[next-1]:
continue
next_num = self.candidates[next]
listnow.append(next_num)
self.dfs(listnow,sumnow+next_num,next+1)
listnow.pop(-1)
def combinationSum2(self, candidates: List[int], target: int) -> List[List[int]]:
self.candidates,self.length,self.target = sorted(candidates),len(candidates),target
self.res,self.vis = [],[False]*self.length
self.dfs([],0,0)
return self.res
def test(data_test):
s = Solution()
data = data_test # normal
# data = [list2node(data_test[0])] # list转node
return s.combinationSum2(*data)
def test_obj(data_test):
result = [None]
obj = Solution(*data_test[1][0])
for fun, data in zip(data_test[0][1::], data_test[1][1::]):
if data:
res = obj.__getattribute__(fun)(*data)
else:
res = obj.__getattribute__(fun)()
result.append(res)
return result
if __name__ == '__main__':
datas = [
[[10,1,2,7,6,1,5],8],
# [[1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1],27],
]
for data_test in datas:
t0 = time.time()
print('-' * 50)
print('input:', data_test)
print('output:', test(data_test))
print(f'use time:{time.time() - t0}s')
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备注:
GitHub:https://github.com/monijuan/leetcode_python
CSDN汇总:模拟卷Leetcode 题解汇总_卷子的博客-CSDN博客
可以加QQ群交流:1092754609
leetcode_python.utils详见汇总页说明
先刷的题,之后用脚本生成的blog,如果有错请留言,我看到了会修改的!谢谢!
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