LeetCode OJ：Combination Sum (组合之和)

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] 

本题需要用到DFS，只不过在在有限搜索的过程中用到了剪枝，使得在优先搜索的过程中一旦遇到了对应的值那么就返回，

不再搜索余下节点。所以这题，首先将数组排序，排序之后再使用DFS加剪枝就可以达到目标。代码如下：

class Solution {
public:
    vector<vector<int>> combinationSum(vector<int>& candidates, int target) {
        tmpCdd = candidates;
        sort(tmpCdd.begin(), tmpCdd.end());
        this->target = target;
        vector<int> tmpVec;
        dfs(0, tmpVec);
        return result;
    }    
private:
    int target;
    vector<int> tmpCdd;
    vector<vector<int>> result;
private:
    void dfs(int index, vector<int> & tmpVec)
    {
        if(index == tmpCdd.size()) return;    //到达叶节点
        int tmpSum = accumulate(tmpVec.begin(), tmpVec.end(), 0);
        if(tmpSum == target){
            result.push_back(tmpVec);
            return;
        }else if(tmpSum > target){//剪枝
            return;
        }else{
            for(int i = index; i < tmpCdd.size(); ++i){//这里从i开始的原因是因为参数可以是重复的
                tmpVec.push_back(tmpCdd[i]);
                dfs(i, tmpVec);
                tmpVec.pop_back();//回溯
            }
        }
    }
};

 java版本的如下所示，思想一样，方法有一点不同，这次不对tmpCdd数组中的值每次都求和，而是每递归一次之后将target的值减去一个数传入 下次递归中，代码如下：

public class Solution {
    List<List<Integer>> ret = new ArrayList<List<Integer>>();
    public List<List<Integer>> combinationSum(int[] candidates, int target) {
        Arrays.sort(candidates);
        for(int i = 0; i < candidates.length; ++i){
            ArrayList<Integer> tmpCdd = new ArrayList<Integer>();
            tmpCdd.add(candidates[i]);
            dfs(i, tmpCdd, candidates, target - candidates[i]);
            tmpCdd.remove(tmpCdd.size() - 1);
        }
        return ret;
    }

    public void dfs(int index, List<Integer> tmpCdd, int[] candidates, int target){
        if(index == candidates.length)
            return;
        if(target < 0)
            return; //直接剪枝
        if(target == 0){
            ret.add(new ArrayList<Integer>(tmpCdd));
            return;
        }else{
            for(int i = index; i < candidates.length; ++i){
                tmpCdd.add(candidates[i]);
                dfs(i, tmpCdd, candidates, target - candidates[i]);
                tmpCdd.remove(tmpCdd.size() - 1);
            }
        }
    }
}