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)