LeetCode OJ

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.

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：

　　设置一个全局数组x来保存每个数使用的数量。x[i] = k 表示第i个数candidate[i]使用了k次。

　　先排序数组，然后使用DFS搜索即可得到解。

代码：

class Solution {
public:
    vector<vector<int>> ans;
    int x[100000];
    void dfs(vector<int> &candidates, int begin, int &cur_sum, int target) {
        if (cur_sum >= target) {
            if (cur_sum == target) {
                vector<int> cur_ans;
                for (int i = 0; i < candidates.size(); i++) {
                    for (int k = 0; k < x[i]; k ++) {
                        
                        cur_ans.push_back(candidates[i]);
                    }
                }
                ans.push_back(cur_ans);
            }
            return;
        }
        else {
            for (int i = begin; i < candidates.size(); i++) {
                x[i] += 1;
                cur_sum += candidates[i];
                dfs(candidates, i, cur_sum, target);
                cur_sum -= candidates[i];
                x[i] -= 1;
            }
        }
        return;
    }
    vector<vector<int> > combinationSum(vector<int> &candidates, int target) {
        for (int i = 0; i < candidates.size(); i++) {
            x[i] = 0;
        }
        int cur_sum = 0;
        sort(candidates.begin(), candidates.end());
        dfs(candidates, 0, cur_sum, target);
        return ans;
    }
};