40. Combination Sum II
https://leetcode.com/problems/combination-sum-ii/
Related problems
39-combination-sum
46-permutations
47-permutations-ii
78-subsets
90-subsets-ii
---
Given a collection of candidate numbers (
candidates) and a target number (target), find all unique combinations in candidates where the candidate numbers sums to target.
Each number in
candidates may only be used once in the combination.
Note:
- All numbers (including
target) will be positive integers. - The solution set must not contain duplicate combinations.
Example 1:
Input: candidates =[10,1,2,7,6,1,5], target =8, A solution set is: [ [1, 7], [1, 2, 5], [2, 6], [1, 1, 6] ]
Example 2:
Input: candidates = [2,5,2,1,2], target = 5, A solution set is: [ [1,2,2], [5] ]---
Related problems
39-combination-sum
46-permutations
47-permutations-ii
78-subsets
90-subsets-ii
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| class Solution { | |
| public List<List<Integer>> combinationSum2(int[] candidates, int target) { | |
| List<List<Integer>> ans = new ArrayList<List<Integer>>(); | |
| Arrays.sort(candidates); | |
| dfs(candidates, 0, target, new ArrayList<Integer>(), ans); | |
| return ans; | |
| } | |
| private void dfs(int[] candidates, int curr, int target, List<Integer> prefix, List<List<Integer>> ans) { | |
| if (target == 0) { | |
| ans.add(new ArrayList<Integer>(prefix)); | |
| return; | |
| } | |
| if (target < 0) { | |
| return; | |
| } | |
| for (int i = curr; i < candidates.length; i++) { | |
| // Skip equal element in the for loop, it will be used in recursed call | |
| if (i > curr && candidates[i] == candidates[i - 1]) { | |
| continue; | |
| } | |
| prefix.add(candidates[i]); | |
| dfs(candidates, i + 1, target - candidates[i], prefix, ans); | |
| prefix.remove(prefix.size() - 1); | |
| } | |
| } | |
| } |
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| class Solution: | |
| def combinationSum2(self, candidates: List[int], target: int) -> List[List[int]]: | |
| def dfs(prefix: List[int], i: int, target: int): | |
| if target == 0: | |
| ans.append(prefix.copy()) | |
| return | |
| for j in range(i, len(candidates)): | |
| if j > i and candidates[j] == candidates[j - 1]: | |
| continue | |
| if target - candidates[j] >= 0: | |
| prefix.append(candidates[j]) | |
| dfs(prefix, j + 1, target - candidates[j]) | |
| prefix.pop() | |
| ans = [] | |
| candidates = sorted(candidates) | |
| dfs([], 0, target) | |
| return ans |
