[LC] 40. Combination Sum 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]
]

Time: O(2^N)
Space: O(N)
class Solution {
    public List<List<Integer>> combinationSum2(int[] candidates, int target) {
        List<List<Integer>> res = new ArrayList<>();
        if (candidates == null || candidates.length == 0) {
            return res;
        }
        Arrays.sort(candidates);
        List<Integer> lst = new ArrayList<>();
        helper(res, lst, candidates, target, 0);
        return res;
    }
    
    private void helper(List<List<Integer>> res, List<Integer> lst, int[] candidates, int reminder, int index) {
        if (reminder < 0) {
            return;
        }
        if (reminder == 0) {
            res.add(new ArrayList<>(lst));
        }
        for (int i = index; i < candidates.length; i++) {

         //i > index will only skip when numbers before cur is used.
        // i > 0 skip all one combination like [1, 1, 6]

            if(i > index && candidates[i] == candidates[i - 1]) {
                continue;
            }
            lst.add(candidates[i]);
            helper(res, lst, candidates, reminder - candidates[i], i + 1);
            lst.remove(lst.size() - 1);
        }
    }
}
原文地址:https://www.cnblogs.com/xuanlu/p/12123410.html