LeetCode 40. 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.
• Elements in a combination (a₁, a₂, … , a_k) must be in non-descending order. (ie, a₁ ≤ a₂ ≤ … ≤ a_k).
• 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]

Almost the same idea as permutation, except that the pos value will have to keep heading forward.

#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;

void combinationSum(vector<int>& candidates, int pos, int& sum, int target, vector< vector<int> >& res, vector<int>& path) { 
    if(sum > target) return; 
    if(sum == target) { 
        res.push_back(path); 
    } 
    for(int i = pos; i < candidates.size(); ++i) { 
        if((i > pos && candidates[i] == candidates[i-1])) continue;  // Draw a graph to show why this line is necessary.
        sum = sum + candidates[i]; 
        path.push_back(candidates[i]); 
        combinationSum(candidates, i + 1, sum, target, res, path); 
        path.pop_back(); 
        sum = sum - candidates[i]; 
    } 
} 
 
// the inputs might have duplicates. 
// the output should be unique sequence. 
vector< vector<int> > combinationSum2(vector<int>& candidates, int target) { 
    if(candidates.size() == 0) return {}; 
    vector< vector<int> > res; 
    vector<int> path; 
    sort(candidates.begin(), candidates.end()); 
    int sum = 0; 
    combinationSum(candidates, 0, sum, target, res, path); 
    return res;
}

int main(void) {
    vector<int> candidates{10, 1, 2, 7, 6, 1, 5};
    vector< vector<int> > res = combinationSum2(candidates, 8);
    for(int i = 0; i < res.size(); ++i) {
        for(int j = 0; j < res[0].size(); ++j) {
            cout << res[i][j] << endl;
        }
        cout << endl;
    }
}