40. 组合总和 II

给定一个候选人编号的集合 candidates 和一个目标数 target ，找出 candidates 中所有可以使数字和为 target 的组合。

candidates 中的每个数字在每个组合中只能使用 一次 。

注意：解集不能包含重复的组合。 

示例 1:

输入: candidates = [10,1,2,7,6,1,5] target = 8
输出:
[
[1,1,6],
[1,2,5],
[1,7],
[2,6]
]

示例 2:

输入: candidates = [2,5,2,1,2] target = 5
输出:
[
[1,2,2],
[5]
]

https://leetcode.cn/problems/combination-sum-ii/description/?envType=company&envId=bytedance&favoriteSlug=bytedance-thirty-days

class Solution {
    List<int[]> freq = new ArrayList<int[]>();
    List<List<Integer>> ans = new ArrayList<List<Integer>>();
    List<Integer> sequence = new ArrayList<Integer>();

    public List<List<Integer>> combinationSum2(int[] candidates, int target) {
        Arrays.sort(candidates);
        for (int num : candidates) {
            int size = freq.size();
            if (freq.isEmpty() || num != freq.get(size - 1)[0]) {
                freq.add(new int[]{num, 1});
            } else {
                ++freq.get(size - 1)[1];
            }
        }
        dfs(0, target);
        return ans;
    }

    public void dfs(int pos, int rest) {
        if (rest == 0) {
            ans.add(new ArrayList<Integer>(sequence));
            return;
        }
        if (pos == freq.size() || rest < freq.get(pos)[0]) {
            return;
        }

        dfs(pos + 1, rest);

        int most = Math.min(rest / freq.get(pos)[0], freq.get(pos)[1]);
        for (int i = 1; i <= most; ++i) {
            sequence.add(freq.get(pos)[0]);
            dfs(pos + 1, rest - i * freq.get(pos)[0]);
        }
        for (int i = 1; i <= most; ++i) {
            sequence.remove(sequence.size() - 1);
        }
    }
}