Combination Sum

Source

• leetcode: 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, 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]

Java

public class Solution {
    public List<List<Integer>> combinationSum(int[] candidates, int target) {
        List<List<Integer>> result = new ArrayList<List<Integer>>();
        Arrays.sort(candidates);
        dfs(candidates, target, 0, new ArrayList<Integer>(), result);
        return result;
    }

    public void dfs(int[] candidates, int target, int pos, List<Integer> path, List<List<Integer>> result){
        if(target<0) return;

        if(target==0){
            result.add(new ArrayList<Integer>(path));
            return;
        }

        for(int i=pos; i<candidates.length; i++){
            path.add(candidates[i]);
            dfs(candidates, target-candidates[i], i, path, result);
            path.remove(path.size()-1);
        }
    }
}

Python

class Solution(object):
    def combinationSum(self, candidates, target):
        """
        :type candidates: List[int]
        :type target: int
        :rtype: List[List[int]]
        """
        result=[]
        candidates.sort()
        self.dfs(candidates, target, 0, [], result)
        return result

    def dfs(self, candidates, target, pos, path, result):
        if target<0:
            return
        if target==0:
            result.append(path[:])
            return

        for i in range(pos, len(candidates)):
            path.append(candidates[i])
            self.dfs(candidates, target-candidates[i], i, path, result)
            path.pop()