Leetcode Combination Sum II

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 is used once in the combination.

Note:

• All numbers (including target) would be positive integers.
• elements in a combination (a 1 ,  a 2 , ...,  a k ) must is in non-descending order. (Ie, a 1  ≤ a 2  ≤ ... ≤ a k ).
• The solution set must not contain duplicate combinations.

For example, given candidate set10,1,2,7,6,1,5and target8,
A Solution set is:
[1, 7]
[1, 2, 5]
[2, 6]
[1, 1, 6]

Test instructions: Or the same as the previous one, the number of combinations, but there is one condition is: Each number can only be used once, can not repeat the result set

Idea: A little change on the line, one is every time the subscript is +1, the other is to avoid repetition, skip the same number

Class Solution {public:    void Dfs (vector<int> candidates, int index, int sum, int target, vector<vector<in T>> &res, vector<int> &path) {        if (sum > Target) return;        if (sum = = target) {            res.push_back (path);            return;        }        for (int i = index; i < candidates.size (); i++) {            path.push_back (candidates[i]);            DFS (candidates, i+1, Sum+candidates[i], target, res, path);            Path.pop_back ();            while (I < candidates.size ()-1 && candidates[i] = = candidates[i+1]) i++;        }    }    vector<vector<int> > CombinationSum2 (vector<int> &candidates, int target) {        sort ( Candidates.begin (), Candidates.end ());        vector<vector<int> > Res;        vector<int> path;        DFS (candidates, 0, 0, Target, res, path);        return res;    }};

Leetcode Combination Sum II