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]
]

AC代码

void digui(vector<int>& candidates, int target, int idx, int sum,
           vector<int>& v, set<vector<int>>& ans) {
    if (sum == target) {
        ans.insert(v);
        return;
    }
    else if (target < sum) return;
    else {
        for (int i = idx + 1; i < candidates.size(); ++i) {
            v.push_back(candidates[i]);
            digui(candidates, target, i, sum + candidates[i], v, ans);
            v.pop_back();
        }
    }
}

class Solution {
public:
    vector<vector<int>> combinationSum2(vector<int>& candidates, int target) {
        sort(candidates.begin(), candidates.end());
        set<vector<int>> ans;
        vector<vector<int>> ret;
        vector<int> v;
        digui(candidates, target, -1, 0, v, ans);
        for (auto t : ans) ret.push_back(t);
        return ret;
    }
};

优化AC代码（运行时去重）

void digui(vector<int>& candidates, int target, int idx, int sum,
           vector<int>& v, vector<vector<int>>& ans) {
    if (sum == target) {
        ans.push_back(v);
        return;
    }
    else if (target < sum) return;
    else {
        for (int i = idx; i < candidates.size(); ++i) {
            if (i != idx && candidates[i] == candidates[i - 1]) continue;
            v.push_back(candidates[i]);
            digui(candidates, target, i + 1, sum + candidates[i], v, ans);
            v.pop_back();
        }
    }
}

class Solution {
public:
    vector<vector<int>> combinationSum2(vector<int>& candidates, int target) {
        sort(candidates.begin(), candidates.end());
        vector<vector<int>> ans;
        vector<int> v;
        digui(candidates, target, 0, 0, v, ans);
        return ans;
    }
};

总结

1、自己写没有考虑去重，上网搜索学习了去重代码，快了整整三倍时间。
2、递归中的for循环，无论是哪一层递归，只要是进入到第二次循环，那么意味着上一个数字已处理完毕，这个时候如果当前数字与上一个数字相同，那么必然会得到相同的结果，有点难理解，想了很久才绕出来，递归的思想还是不熟。