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.

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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]
]
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Input: candidates = [2,5,2,1,2], target = 5,
A solution set is:
[
  [1,2,2],
  [5]
]

给定的candidates中可能有duplicates, 所以sort并if (i > start && nums[i] == nums[i-1]) continue;去掉duplicates;

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class Solution {
    public List<List<Integer>> combinationSum2(int[] nums, int target) {
        List<List<Integer>> list = new ArrayList<>();
        Arrays.sort(nums);
        dfs(list, new ArrayList<>(), nums, target, 0);
        return list;
    }
    
    public void dfs(List<List<Integer>> list, List<Integer> arr, int[] nums, int target, int start) {
        if (target < 0) return;
        else if (target == 0) list.add(new ArrayList<>(arr));
        else {
            for (int i = start; i < nums.length; ++i) {
                if (i > start && nums[i] == nums[i-1]) continue;
                arr.add(nums[i]);
                dfs(list, arr, nums, target - nums[i], i + 1); // i, can be reused
                arr.remove(arr.size() - 1);
            }
        }
    }
}