Given an array of distinct integers candidates and a target integer target, return a list of all unique combinations of candidates where the chosen numbers sum to target. You may return the combinations in any order. The same number may be chosen from candidates an unlimited number of times. Two combinations are unique if the frequency of at least one of the chosen numbers is different. It is guaranteed that the number of unique combinations that sum up to target is less than 150 combinations for the given input.
Example 1: Input: candidates = [2,3,6,7], target = 7 Output: [[2,2,3],[7]] Explanation: 2 and 3 are candidates, and 2 + 2 + 3 = 7. Note that 2 can be used multiple times. 7 is a candidate, and 7 = 7. These are the only two combinations. Example 2: Input: candidates = [2,3,5], target = 8 Output: [[2,2,2,2],[2,3,3],[3,5]]
- code dp
class Solution:
def combinationSum(self, candidates: List[int], target: int) -> List[List[int]]:
dp = [[] for _ in range(target + 1)]
dp[0] = [[]]
for candidate in candidates:
for index in range(candidate, target + 1):
for combo in dp[index - candidate]:
dp[index].append(combo + [candidate])
return dp[target]
- code dfs
class Solution:
def combinationSum(self, candidates: List[int], target: int) -> List[List[int]]:
res = []
def dfs(start, target, path):
if target < 0:
return
elif target == 0:
res.append(path)
return
for i in range(start, len(candidates)):
dfs(i, target - candidates[i], path + [candidates[i]])
dfs(0, target, [])
return res
- code bfs
class Solution:
def combinationSum(self, candidates: List[int], target: int) -> List[List[int]]:
q = collections.deque([(i, [c]) for i, c in enumerate(candidates)])
res = []
while q:
start, cur = q.popleft()
if sum(cur) == target:
res.append(cur)
for i in range(start, len(candidates)):
if sum(cur + [candidates[i]]) <= target:
q.append((i, cur + [candidates[i]]))
return res