39. Combination Sum 解题记录

题目描述：

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]
]

题目给我们一堆数字（没有重复），要我们找到所有和为target的组合（不能重复），每个数字能多次使用。

解题思路：

这题我想的是用递归函数将所有情况都试一遍。

代码：

class Solution {
public:
    vector<vector<int>> res;
    void selectAll(vector<int>& candidates, vector<int> nums, int target, int index){
     //candidates为题目所给数组，nums是进行存储数组，target是目标数字（每一层都会变），index是记录上一层nums存储的数字在candidates中的位置（防止重复一维数组出现）
        if(target == 0){
        //target为0时存一维数组，小于0结束
            res.push_back(nums);
        }
        else if(target > 0){
            int n = candidates.size();
            for(int i = index; i < n; i++){
            //进行每一层每种情况的遍历
                vector<int> tempNums(nums);
                tempNums.push_back(candidates[i]);
                selectAll(candidates, tempNums, target-candidates[i], i);
            }
        }
    }
    vector<vector<int>> combinationSum(vector<int>& candidates, int target) {
        vector<int> nums;
        selectAll(candidates, nums, target, 0);
        return res;
    }
};