40. Combination Sum II

• Difficulty: Medium

• Topics: Array, Backtracking

• Similar Questions:

  • Combination Sum

Problem:

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]
]

Solutions:

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

    void helper(vector<int>& candidates, int pos, int target, vector<int>& path, vector<vector<int>>& ret) {
        if (target < 0) return;
        if (pos == candidates.size()) {
            if (target == 0) {
                ret.push_back(path);
            }
            return;
        }

        int val = candidates[pos];
        path.push_back(val);
        helper(candidates, pos + 1, target - val, path, ret);
        path.pop_back();
        while (pos + 1 < candidates.size() && candidates[pos + 1] == candidates[pos]) ++pos;
        helper(candidates, pos + 1, target, path, ret);
    }
};