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, … ,
ak) must be in non-descending order. (ie, a1 ≤
a2 ≤ … ≤ ak).
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;
}
}