39. Combination Sum

Given a set of candidate numbers (C) (without duplicates) 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.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]
]

给定一个无重复数值的集合以及一个目标数值，在集合中找到所有和为目标数值的子集合。

#include <algorithm>#include "CombinationSum.h"vector<vector<int>> CombinationSum::combinationSum(vector<int>& candidates, int target){    sort(candidates.begin(), candidates.end());    vector<std::vector<int> > res;    vector<int> combination;    //开始迭代    combinationSum(candidates, target, res, combination, 0);    return res;}void CombinationSum::combinationSum(std::vector<int> &candidates, int target, std::vector<std::vector<int> > &res, std::vector<int> &combination, int begin) {    if (!target) {        res.push_back(combination);        return;    }    for (int i = begin; i != candidates.size() && target >= candidates[i]; ++i) {        combination.push_back(candidates[i]);        //每次查找的目标元素不同        combinationSum(candidates, target - candidates[i], res, combination, i);        //这步保证遍历完所有子集        combination.pop_back();    }}