标签：find 重复 number from dfs ons div repeat color
题目：
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.
Example
Given candidate set [2,3,6,7] and target 7, a solution set is:
[7]
[2, 2, 3]题解：
这个题和LeetCode不一样，这个允许重复数字产生，那么输入[2,2,3],7 结果为[2, 2, 3], [2, 2, 3], [2, 2, 3];要么对最后结果去除重复数组，要么在处理之前就对candidates数组去重复，或者利用set不重复的特点添加数组。
Solution 1 ()
class Solution { public: vector<vector<int> > combinationSum(vector<int> &candidates, int target) { if (candidates.empty()) { return {{}}; } set<vector<int> > res; vector<int> cur; sort(candidates.begin(), candidates.end()); dfs(res, cur, candidates, target, 0); return vector<vector<int> > (res.begin(), res.end()); } void dfs(set<vector<int> > &res, vector<int> &cur, vector<int> candidates, int target, int pos) { if (!cur.empty() && target == 0) { res.insert(cur); return; } for (int i = pos; i < candidates.size(); ++i) { if (target - candidates[i] >= 0) { cur.push_back(candidates[i]); dfs(res, cur, candidates, target - candidates[i], i); cur.pop_back(); } else { break; } } } };
标签：find 重复 number from dfs ons div repeat color
原文地址：http://www.cnblogs.com/Atanisi/p/6869739.html
