Combination Sum II

Description

Given a collection of candidate numbers (candidates) and a target number (target), find all unique combinations in candidates where the candidate numbers sum to target.

Each number in candidates may only be used once in the combination.

Note: The solution set must not contain duplicate combinations.

Example 1:

Input: candidates = [10,1,2,7,6,1,5], target = 8 Output: [ [1,1,6], [1,2,5], [1,7], [2,6] ]

Example 2:

Input: candidates = [2,5,2,1,2], target = 5 Output: [ [1,2,2], [5] ]

Constraints:

• 1 <= candidates.length <= 100
• 1 <= candidates[i] <= 50
• 1 <= target <= 30

Code

延續 Combination Sum 的思路，只是將 helper 呼叫的 start 設為 i + 1 而不是 i ，因為 candidate 不能被 reuse。

另外使用在 Permutations II 中學到的技巧去避免 duplicate。

if(used[i] || i > 0 && candidates[i] == candidates[i-1] && !used[i-1]) continue;

class Solution {
public:
    vector<vector<int>> combinationSum2(vector<int>& candidates, int target) {
        vector<vector<int>> answer;
        vector<int> curr;
        sort(candidates.begin(), candidates.end());
        int n = candidates.size();
        vector<bool> used(n, false);
        helper(answer, candidates, curr, 0, 0, target, used);
        return answer;
    }
 
    void helper(vector<vector<int>> &answer, vector<int> &candidates, vector<int> curr, int currSum, int start, int target, vector<bool> used) {
 
        if(currSum > target) return;
        if(currSum == target) {
            answer.push_back(curr);
            return;
        }
 
        for(int i = start; i < candidates.size(); i++) {
            // avoid duplicate
            if(used[i] || i > 0 && candidates[i] == candidates[i-1] && !used[i-1]) continue;
            currSum += candidates[i];
            used[i] = true;
            curr.push_back(candidates[i]);
            // set start to i + 1, since elements cannot be reuse
            helper(answer, candidates, curr, currSum, i + 1, target, used);
            currSum -= candidates[i];
            used[i] = false;
            curr.pop_back();
        }
    }
};

Source

• Combination Sum II - LeetCode