2014
11-18

# LeetCode-Combination Sum[DFS]

### Combination Sum

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, a1a2 ≤ … ≤ 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]

// LeetCode, Combination Sum
// 时间复杂度O(n!)，空间复杂度O(n)
class Solution {
public:
vector<vector<int> > combinationSum(vector<int> &nums, int target) {
sort(nums.begin(), nums.end());
vector<vector<int> > result; // 最终结果
vector<int> intermediate; // 中间结果
dfs(nums, target, 0, intermediate, result);
return result;
}

private:
void dfs(vector<int>& nums, int gap, int start, vector<int>& intermediate,
vector<vector<int> > &result) {
if (gap == 0) {  // 找到一个合法解
result.push_back(intermediate);
return;
}
for (size_t i = start; i < nums.size(); i++) { // 扩展状态
if (gap < nums[i]) return; // 剪枝

intermediate.push_back(nums[i]); // 执行扩展动作
dfs(nums, gap - nums[i], i, intermediate, result);
intermediate.pop_back();  // 撤销动作
}
}
};


Java代码:

public class Solution {
public static List<List<Integer>> combinationSum(int[] candidates, int target) {
List<List<Integer>> res = new ArrayList<List<Integer>>();
List<Integer> list = new ArrayList<Integer>();

Arrays.sort(candidates);
combinationSumRecur(candidates,0, target, list, res);
//System.out.println(res.size());
//System.out.println(res);
return res;
}

public  static void combinationSumRecur(int candidates[],int k,int target,List<Integer> list,List<List<Integer>> res){
if(target < 0 || k >= candidates.length) return;
if(0 == target){
}