康瑞部落

Combination Sum II

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.

• 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 10,1,2,7,6,1,5 and target 8, 
A solution set is: 
[1, 7] 
[1, 2, 5] 
[2, 6] 
[1, 1, 6] 

解题思路：

与combination sum类似，与之不同的是，每个候选元素只能用一次。

class Solution {
public:
    vector<vector<int>> combinationSum2(vector<int>& candidates, int target) {
        vector<vector<int>> result;
        int len=candidates.size();
        if(len==0){
            return result;
        }
        std:sort(candidates.begin(), candidates.end());
        map<int, int> keyToNumber;  //相当于系数,表示每个数出现了多少次，注意，这里变成下标，而非数了。
        set<vector<int>> contains;  //是否包含了这组解
        getResult(result, contains, candidates, 0, keyToNumber, target);
    }
    
    void getResult(vector<vector<int>>& result, set<vector<int>>& contains, vector<int>& uniqueCandidates, int candidateIndex, map<int, int>& keyToNumber, int left){
        if(left<0){
            return;
        }
        if(left==0){
            vector<int> item;
            for(map<int, int>::iterator it=keyToNumber.begin(); it!=keyToNumber.end(); it++){
                int number=it->second;
                int key = it->first;
                for(int i=0; i<number; i++){
                    item.push_back(uniqueCandidates[key]);
                }
            }
            if(contains.find(item)==contains.end()){
                result.push_back(item);
                contains.insert(item);
            }
            return;
        }
        if(candidateIndex>=uniqueCandidates.size()){
            return;
        }
        int number=0;
        while(left>=0&&number<2){   //用0次或一次
            if(number!=0)
                keyToNumber[candidateIndex]=number;
            getResult(result, contains, uniqueCandidates, candidateIndex+1, keyToNumber, left);
            if(number!=0){
                keyToNumber.erase(candidateIndex);
            }
            left = left-uniqueCandidates[candidateIndex];
            number++;
        }
    }
};

二次刷题（2015-08-18）

class Solution {
public:
    vector<vector<int>> combinationSum2(vector<int>& candidates, int target) {
        std::sort(candidates.begin(), candidates.end());
        vector<vector<int>> result;
        helper(candidates, vector<int>(), 0, target, result);
        return result;
    }
    
    void helper(vector<int>& candidates, vector<int> item, int start, int left, vector<vector<int>>& result){
        if(left == 0){
            if(item.size() != 0){
                result.push_back(item);
            }
            return;
        }
        if(candidates[start] > left){
            return;
        }
        for(int i = start; i<candidates.size(); i++){
            if(i>start && candidates[i]==candidates[i-1]){    //这里防止重复
                continue;
            }
            if(left>=candidates[i]){
                item.push_back(candidates[i]);
                helper(candidates, item, i + 1, left - candidates[i], result);    //这里的i+1防止一个元素选多次
                item.pop_back();
            }else{
                break;
            }
        }
    }
};