[LeetCode] Combinations Sum

문제

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.

예제

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]]

Example 3:

Input: candidates = [2], target = 1
Output: []

Example 4:

Input: candidates = [1], target = 1
Output: [[1]]

Example 5:

Input: candidates = [1], target = 2
Output: [[1,1]]

조건

• 1 <= candidates.length <= 30
• 1 <= candidates[i] <= 200
• All elements of candidates are distinct.
• 1 <= target <= 500

답

풀이 1

BFS를 활용한 풀이

class Solution:
    def combinationSum(self, candidates: List[int], target: int) -> List[List[int]]:
        q = collections.deque()
        ans = []
        for c in candidates:
            q.append([c,[c]])
        
        while q:
            num, li = q.popleft()
            
            if num > target:
                continue
            elif num == target:
                li.sort()
                if not li in ans:
                    ans.append(li)
                continue
            
            for i in candidates:
                q.append([num+i,li+[i]])
        
        return ans

풀이 2

DFS를 활용한 풀이

class Solution:
    def combinationSum(self, candidates: List[int], target: int) -> List[List[int]]:
        result = []
        def dfs(csum, index, path):
            if csum < 0:
                return
            if csum == 0:
                result.append(path)
                return
            
            for i in range(index, len(candidates)):
                dfs(csum-candidates[i], i, path+[candidates[i]])
                
        dfs(target, 0, [])
        return result