Leetcode 39. Combination Sum (递归)

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]
]

题目链接：https://leetcode.com/problems/combination-sum/description/

题目大意：给定一个数组和一个目标数字。求数组中和为目标数字的所有数字组合，数字可以重复相加。

解题思路：dfs递归，到最底层若满足条件就把数字组合加入答案集合。

代码如下：

class Solution {
public:
    vector<vector<int> > ans;
    vector<int> ca;
    int n;
    void dfs(int st,vector<int> a,int cur,int target){  
        if(cur == target){
            ans.push_back(a);  //a是每个可能的答案，如果最终满足条件，即a中所有数字和为target，则把a加入答案数组
            return;
        }
        if(st >= n)
            return;
        for(int i = st;i < n;i++){
            if(cur + ca[i] <= target){
                int num = 1;
                vector <int> b = a;
                while(cur + ca[i] * num <= target){   //数字可以重复，多次加入同一个数字
                    b.push_back(ca[i]);
                    dfs(i + 1,b,cur + ca[i] * num,target);  //i是起始位置 
                    num ++;
                }
            }
        }
    }
    vector<vector<int> > combinationSum(vector<int>& candidates, int target) {
        n = candidates.size();
        ca = candidates;
        vector<int> a;
        sort(ca.begin(),ca.end());   //排序
        unique(ca.begin(), ca.end());   //去重
        dfs(0,a,0,target);  
        return ans;
    }
};