動機

複習backtrack backtrack也是遞迴，所以也是看有什麼case(可以做的動作)

Problem

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 = 7Output: [[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 = 8Output: [[2,2,2,2],[2,3,3],[3,5]]

Example 3:

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

Example 4:

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

Example 5:

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

 

Constraints:

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

Sol

一次把這個數字用完，不論是不選、選一次、選好幾次都在這層一次做完

def dfs(cs,t,i,tup,acc):
    if len(cs) <= i:
        return [tup] if t == acc else []
    else:
        ret = dfs(cs,t,i+1,tup,acc)
        while t > acc:
            acc += cs[i]
            tmp.append(cs[i])
            ret += dfs(cs,t,i+1,tmp,acc)
        return ret
    
class Solution:
    def combinationSum(self, cs: List[int], t: int) -> List[List[int]]:
        return dfs(cs,t,0,[],0)