39. Combination Sum**

​​https://leetcode.com/problems/combination-sum/​​

题目描述

Given a set of candidate numbers (​​candidates​​) (without duplicates) and a ​​target​​​ number (target), find all unique combinations in ​​candidates​​​ where the candidate numbers sums to ​​target​​.

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

Note:

All numbers (including ​​target​​​) will be positive integers.
The solution set must not contain duplicate combinations.
Example 1:

Input: candidates = [2,3,6,7], target = 7,
A solution set is:
[
  [7],
  [2,2,3]
]

Example 2:

Input: candidates = [2,3,5], target = 8,
A solution set is:
[
  [2,2,2,2],
  [2,3,3],
  [3,5]
]

C++ 实现 1

DFS + Backtracing. 注意数组中是没有重复数字的, 因此降低了难度, 估计其他类似的题会逐渐增大难度. 另外, 要保证 ​​res​​​ 中的结果是不重复的, 那么还需给 ​​dfs​​​ 函数增加 ​​start​​​ 参数, 每次 ​​for​​​ 循环都需要从 ​​i = start​​ 开始.

class Solution {
private:
    void dfs(const vector<int>& nums, int target, int start, vector<int> &cur, vector<vector<int>> &res) {
        if (target == 0) {
            res.push_back(cur);
            return;
        }
        for (int i = start; i < nums.size() && nums[i] <= target; ++ i) {
            cur.push_back(nums[i]);
            dfs(nums, target - nums[i], i, cur, res);
            cur.pop_back();
        }
    }
public:
    vector<vector<int>> combinationSum(vector<int>& candidates, int target) {
        std::sort(candidates.begin(), candidates.end());
        vector<vector<int>> res;
        vector<int> cur;
        dfs(candidates, target, 0, cur, res);
        return res;
    }
};