Combination Sum II, III -- LeetCode 40, 216

[Questions]
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 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.

For example, given candidate set [10, 1, 2, 7, 6, 1, 5] and target 8, 
A solution set is: 

[
  [1, 7],
  [1, 2, 5],
  [2, 6],
  [1, 1, 6]
]

Combination Sum III:

Find all possible combinations of k numbers that add up to a number n, given that only numbers from 1 to 9 can be used and each combination should be a unique set of numbers.

Example 1:

Input: k = 3, n = 7

Output:

[[1,2,4]]

Example 2:

Input: k = 3, n = 9

Output:

[[1,2,6], [1,3,5], [2,3,4]]

[Thoughts]
Similar to "Combination Sum", those are just variations. The same code skeleton of backtracking on a vector can be reused. The variations are marked in the following code.

[Solution]
//-- Combination Sum II --
class Solution {
    void backtrack(vector<int>& cand, int target, int start, vector<int>& tmplist, vector<vector<int>>& res) {
        if (target<0) return;
        if (target==0) {
            res.push_back(tmplist); return;
        }

        for (int i=start; i<cand.size();i++) {

            tmplist.push_back(cand[i]);
            backtrack(cand, target-cand[i], i+1, tmplist, res);
            tmplist.pop_back();
            
            while(cand[i+1]==cand[i]) i++; //skip duplicates
        }
    }
public:
    vector<vector<int>> combinationSum2(vector<int>& candidates, int target) {
        vector<vector<int>> res;
        vector<int> sublist;
        sort(candidates.begin(), candidates.end());
        backtrack(candidates, target, 0, sublist, res);
        return res;
    }
};

//-- Combination Sum III --
class Solution {
    void backtrack(int len, int target, int start, vector<int>&tmplist, vector<vector<int>>& res) {
        if (target<0) return;
        if (target==0 && tmplist.size()==len) {
            res.push_back(tmplist); return;
        }
        if (tmplist.size()>len) return;  // length limit
        
        for (int i=start; i<10; i++) {
            tmplist.push_back(i);
            backtrack(len, target-i, i+1, tmplist, res);
            tmplist.pop_back();
        } 
    }
public:
    vector<vector<int>> combinationSum3(int k, int n) {
        vector<vector<int>> res;
        vector<int> tmplist;
        backtrack(k, n, 1, tmplist, res);
        return res;
    }

};