40.Combination Sum II

Tags: Medium Backtracking Array

Links: https://leetcode.com/problems/combination-sum-ii/

---

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]
]

---

class Solution {
    void DFS(vector<int>& candidates, int gap, int startPosition, vector<int> &tmp, vector<vector<int>> &result)
    {
        if (gap == 0){
            result.push_back(tmp);
            return;
        }

        for (size_t i = startPosition; i < candidates.size(); ++i){
            if (gap < candidates[i]) return; 

            tmp.push_back(candidates[i]);
            DFS(candidates, gap - candidates[i], i + 1, tmp, result);
            tmp.pop_back();
        }
    }

public:
    vector<vector<int>> combinationSum2(vector<int>& candidates, int target) {
        vector<vector<int>> result;
        vector<int> tmp;

        sort(candidates.begin(), candidates.end());

        DFS(candidates, target, 0, tmp, result);
        sort(result.begin(), result.end());
        result.erase(unique(result.begin(), result.end()), result.end());

        return result;
    }
};

其实这道题和3Sum很接近，或者K-Sum的问题，需要去掉重复元素的计算。

class Solution {
    void DFS(vector<int>& candidates, int gap, int startPosition, vector<int> &tmp, vector<vector<int>> &result)
    {
        if (gap == 0){
            result.push_back(tmp);
            return;
        }

        int pre = -1;
        for (size_t i = startPosition; i < candidates.size(); ++i){
            if (gap < candidates[i]) return; 

            if (candidates[i] == pre) continue;
            pre = candidates[i];
            tmp.push_back(candidates[i]);
            DFS(candidates, gap - candidates[i], i + 1, tmp, result);
            tmp.pop_back();
        }
    }

public:
    vector<vector<int>> combinationSum2(vector<int>& candidates, int target) {
        vector<vector<int>> result;
        vector<int> tmp;

        sort(candidates.begin(), candidates.end());

        DFS(candidates, target, 0, tmp, result);

        return result;
    }
};