[LeetCode] 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]
]

题意：给一个数组，找出所有列表满足，列表和等于target
题不难，回溯算法，大家注意一点就行了，有重复的元素，使用set

class Solution {
    public List> combinationSum2(int[] candidates, int target) {
        Set> res = new HashSet<>();
        if (candidates.length == 0)
            return new ArrayList<>(res);
        Arrays.sort(candidates);
        DFS(res, new ArrayList(), candidates, target, 0, 0);
        return new ArrayList<>(res);
    }

    private void DFS(Set> res, ArrayList list, int[] candidates, int target, int index, int sum) {
        for (int i = index; i < candidates.length; i++) {
            sum += candidates[i];
            list.add(candidates[i]);
            if (sum == target)
                res.add(new ArrayList<>(list));
            else if (sum < target)
                DFS(res, list, candidates, target, i + 1, sum);
            else if (sum > target) {
                sum -= candidates[i];
                list.remove(list.size() - 1);
                break;
            }
            sum -= candidates[i];
            list.remove(list.size() - 1);
        }
    }
}

 

转载于:https://www.cnblogs.com/Moriarty-cx/p/9942971.html