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]
]
题意：
给定一个集合以及一个值target，找出所有加起来等于target的组合。（每个元素只能用一次）
Solution1： Backtracking
在[leetcode]39. Combination Sum组合之和的基础上， 加一行查重的动作。
code
class Solution {
public List<List<Integer>> combinationSum2(int[] candidates, int target) {
Arrays.sort(candidates);
List<List<Integer>> result = new ArrayList<>();
List<Integer> path = new ArrayList<>();
helper(candidates, 0, target, path, result );
return result;
}
private void helper(int[] nums, int index, int remain, List<Integer> path, List<List<Integer>> result){
if (remain == 0){
result.add(new ArrayList<>(path));
return;
}
for(int i = index; i < nums.length; i++){
if (remain < nums[i]) return;
if(i > index && nums[i] == nums[i-1]) continue; /** skip duplicates */
path.add(nums[i]);
helper(nums, i + 1, remain - nums[i], path, result);
path.remove(path.size() - 1);
}
}
}