0040. 组合总和 II [Medium] [Combination Sum II].md

40. 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]
]

---

题目: 给定一个候选元素的数组candidates，找出满足和为target的所有子集。candidates可能包含重复的元素，只能使用candidates中存在的元素。

思路: 和4sum类似，只是这里的递归深度不定，递归返回的条件是remain已经等于0，参考A general approach to backtracking questions。同时由于题目的特殊性，candidates中的元素和target均为正数，所以当remain小于0时也可以直接返回，减少递归的次数。为了防止结果中出现重复的集合，先对candidates排序，同时在每层递归中跳过已经判断过的相同元素。

工程代码下载

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

class Solution {
public:
    vector<vector<int>> combinationSum2(vector<int>& candidates, int target) {
        sort(candidates.begin(), candidates.end());
        vector<vector<int> > res;
        vector<int> vec;
        helper(candidates, res, vec, 0, target);
        return res;
    }
private:
    void helper(vector<int>& candidates, vector<vector<int>>& res, vector<int>& vec, int start, int remain){
        /** All numbers (including target) will be positive integers. **/
        if(remain < 0)
            return;

        if(remain == 0){
            res.push_back(vec);
            return;
        }

        for(int i = start; i< candidates.size(); ++i){
            /** skip duplicates */
            if(i > start && candidates[i] == candidate[i-1])
                continue;
            vec.push_back(candidates[i]);
            helper(candidates, res, vec, i+1, remain-candidates[i]);
            vec.pop_back(); 
        }
    }
};

int main(int argc, char const *argv[])
{
    Solution sln;
    vector<int> testcase{10,1,2,7,6,1,5};
    vector<vector<int>> res = sln.combinationSum2(testcase, 8);
    for(auto vec: res){
        for(auto i : vec)
            cout << i << ",";
        cout << endl;
    }
    return 0;
}