LeetCode | 40. Combination Sum II

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本文介绍了一种解决组合求和问题的算法实现方案。针对给定候选数集合与目标数，寻找所有不重复的组合使得候选数之和等于目标数。通过递归方法结合剪枝策略，在保证组合唯一性的基础上提高了算法效率。

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.
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]
]

无重复的组合，数组中每个元素只能使用一次。和 Combination Sum I 类似，只要添加一些限定条件即可。

//19ms
class Solution {
public:
    void f(vector<int> &candidates,int target,vector<vector<int> > &res,vector<int> &tmp,int _index)
    {
        if(target < 0)
            return;
        if(target == 0)
        {
            int flag = 0;   //flag = 1代表有重复
            for(int i=0;i<res.size();i++)
            {
                if(res[i].size() == tmp.size())
                {
                    for(int j=0;j<tmp.size();j++)
                    {
                        flag = 1;
                        if(res[i][j] != tmp[j])
                        {
                            flag = 0; break;
                        }
                    }
                }
                if(flag == 1)
                    break;
            }
            if(flag == 0)
                res.push_back(tmp);

            return;
        }
        for(int i=_index;i<candidates.size();i++)
        {
            tmp.push_back(candidates[i]);
            f(candidates,target-candidates[i],res,tmp,i+1);
            tmp.pop_back();
        }
    }

    vector<vector<int> > combinationSum2(vector<int>& candidates, int target)
    {
        sort(candidates.begin(),candidates.end());
        vector<vector<int> > res;
        vector<int> tmp;
        f(candidates,target,res,tmp,0);
        return res;
    }
};

*这是网上提供的解法，不用每次遍历已有的 res，直接判重。性能提升飞快

//6ms
class Solution {
public:
    vector<vector<int> > combinationSum2(vector<int> &num, int target) 
    {
        vector<vector<int>> res;
        sort(num.begin(),num.end());
        vector<int> local;
        findCombination(res, 0, target, local, num);
        return res;
    }
    void findCombination(vector<vector<int>>& res, const int order, const int target, vector<int>& local, const vector<int>& num)
    {
        if(target==0)
        {
            res.push_back(local);
            return;
        }
        else
        {
            for(int i = order;i<num.size();i++) // iterative component
            {
                if(num[i]>target) return;
                if(i&&num[i]==num[i-1]&&i>order) continue; // check duplicate combination
                local.push_back(num[i]),
                findCombination(res,i+1,target-num[i],local,num); // recursive componenet
                local.pop_back();
            }
        }
    }
};