leetcode解题: Combination Sum (39)

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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:

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[
  [7],
  [2, 2, 3]
]

解法1：

经典的backtracking的做法。 一个可能的小改进是可以先对原数组进行排序，然后再判断target是否还大于当前可选的节点，这样可以提高一点效率。
C++

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class Solution {
public:
    vector<vector<int>> combinationSum(vector<int>& candidates, int target) {
        vector<vector<int>> res;
        if (candidates.size() == 0 || target <= 0) {
            return res;
        }
        vector<int> cur;
        
        helper(candidates, target, 0, cur, res);
        return res;
    }
    
    void helper(vector<int>& candidates, int target, int pos, vector<int>& cur, vector<vector<int>>& res) {
        if (target < 0) {
            return;
        }
        if (target == 0) {
            res.push_back(cur);
            return;
        }
        
        for (int i = pos; i < candidates.size(); ++i) {
            cur.push_back(candidates[i]);
            helper(candidates, target - candidates[i], i,cur, res);
            cur.pop_back();
        }
        return;
    }
};

Java

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