40 Combination Sum II

40. Combination Sum II

1. Question

Given a collection of candidate numbers (C) and a target number (T), find all unique combinations in C where the candidate numbers sums to T.

Each number in C 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.

For example, given candidate set[10, 1, 2, 7, 6, 1, 5]and target8, A solution set is:

[
  [1, 7],
  [1, 2, 5],
  [2, 6],
  [1, 1, 6]
]

2. Implementation

(1) Backtracking

class Solution {
    public List<List<Integer>> combinationSum2(int[] candidates, int target) {
        List<List<Integer>> res = new ArrayList<>();
        List<Integer> combinations = new ArrayList<>();
        Arrays.sort(candidates);
        getCombinations(candidates, 0, 0, target, combinations, res);
        return res;
    }

    public void getCombinations(int[] candidates, int index, int sum, int target, List<Integer> combinations, List<List<Integer>> res) {
        if (sum > target) {
            return;
        }

        if (sum == target) {
            res.add(new ArrayList<>(combinations));
            return;
        }

        for (int i = index; i < candidates.length; i++) {
            if (i > index && candidates[i - 1] == candidates[i]) {
                continue;
            }

            combinations.add(candidates[i]);
            getCombinations(candidates, i + 1, sum + candidates[i], target, combinations, res);
            combinations.remove(combinations.size() - 1);
        }
    }
}

3. Time & Space Complexity

Backtracking: 时间复杂度O(2^n), 空间复杂度O(2^n)