문제 설명
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]]
Example 3:
Input: candidates = [2], target = 1
Output: []
문제 풀이
class Solution {
public List<List<Integer>> combinationSum(int[] candidates, int target) {
List<List<Integer>> answer = new ArrayList<>();
backtracking(0, target, candidates.length, answer, new ArrayList<Integer>(), candidates);
return answer;
}
public void backtracking(int start, int target, int n, List<List<Integer>> answer,
List<Integer> cur, int[] candidates) {
if(target == 0) {
answer.add(new ArrayList<>(cur));
} else if(target < 0) {
return;
} else {
for(int i = start; i < n; i++) {
cur.add(candidates[i]);
backtracking(i, target - candidates[i], n, answer, cur, candidates);
cur.remove(cur.size() - 1);
}
}
}
}일반 combination 문제와 다른점
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중복이 허용된다. 그렇기 때문에 재귀적으로 들어갈때 for loop의 시작인덱스를 증가시키지 않는다.
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순서가 상관없는 모든 경우의 수를 전부 탐색하기 때문에 길이를 정하지 않는다. 대신 target이 0이 되는경우를 base case로 잡는다.
