Combination Sum

Given a set of candidate numbers (C) and a target number (T), find all unique combinations in C where the candidate numbers sums to T.

The same repeated number may be chosen from C unlimited number of times.

Note:

• All numbers (including target) will be positive integers.

• Elements in a combination (a1, a2, … , ak) must be in non-descending order. (ie, a1 ≤ a2 ≤ … ≤ ak).

• The solution set must not contain duplicate combinations.

For example, given candidate set 2,3,6,7 and target 7, 
A solution set is: 
[7] 
[2, 2, 3] 

解题思路：

这道题的题意是从n个正整数中选出值为特定值的所有元素，这n个数中每个数可以重复选用。

这是一个np难问题，暴力法。主要是通过回溯暴力。代码如下：

class Solution {
public:
    vector> combinationSum(vector& candidates, int target) {
        vector> result;
        int len=candidates.size();
        if(len==0){
            return result;
        }
        std:sort(candidates.begin(), candidates.end());
        map keyToNumber;  //相当于系数,表示每个数出现了多少次
        getResult(result, candidates, 0, keyToNumber, target);
    }
    
    void getResult(vector>& result, vector& uniqueCandidates, int candidateIndex, map& keyToNumber, int left){
        if(left<0){
            return;
        }
        if(left==0){
            vector item;
            for(map::iterator it=keyToNumber.begin(); it!=keyToNumber.end(); it++){
                int number=it->second;
                int key = it->first;
                for(int i=0; i=uniqueCandidates.size()){
            return;
        }
        int number=0;
        while(left>=0){
            if(number!=0)
                keyToNumber[uniqueCandidates[candidateIndex]]=number;
            getResult(result, uniqueCandidates, candidateIndex+1, keyToNumber, left);
            if(number!=0){
                keyToNumber.erase(uniqueCandidates[candidateIndex]);
            }
            left = left-uniqueCandidates[candidateIndex];
            number++;
        }
    }
};

二次刷题（2015-08-18）

class Solution {
public:
    vector> combinationSum(vector& candidates, int target) {
        std::sort(candidates.begin(), candidates.end());
        vector> result;
        helper(candidates, vector(), 0, target, result);
        return result;
    }
    
    void helper(vector& candidates, vector item, int start, int left, vector>& result){
        if(left == 0){
            if(item.size() != 0){
                result.push_back(item);
            }
            return;
        }
        if(candidates[start] > left){
            return;
        }
        for(int i = start; i=candidates[i]){
                item.push_back(candidates[i]);
                helper(candidates, item, i, left - candidates[i], result);
                item.pop_back();
            }else{
                break;
            }
        }
    }
};

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