10.7 Combination Sum

Link: https://oj.leetcode.com/problems/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] 

My first attempt: (the code is wrong). 

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

Correct Solution:

Time: O(n!), Space: O(n)

public class Solution {    public ArrayList<ArrayList<Integer>> combinationSum(int[] candidates, int target) {        ArrayList<ArrayList<Integer>> result = new ArrayList<ArrayList<Integer>>();        Arrays.sort(candidates);        ArrayList<Integer> item = new ArrayList<Integer>();        dfs(candidates, 0, item, target, result);        return result;    }        public void dfs(int[] candidates, int start, ArrayList<Integer> item, int target, ArrayList<ArrayList<Integer>> result){        if(target == 0){            result.add(new ArrayList<Integer>(item));        }        for(int i = start; i < candidates.length; i++){            if(target < candidates[i]) return;            if(i > 0 && candidates[i] == candidates[i-1]) continue;//因为每个数本身就可以重复取，所以数组里的相同数可以跳过。            item.add(candidates[i]);            dfs(candidates, i, item, target-candidates[i], result);//dfs里的参数还是i,而不是i+1是因为每个数可以重复取。            item.remove(item.size()-1);        }    }}

I think adding this line can cut off paths in dfs search. 

if(target < candidates[i]) return;

But if we don't have this line, we need to check if(target < 0):

 public void dfs(int[] candidates, int start, ArrayList<Integer> item, int target, ArrayList<ArrayList<Integer>> result){        if(target < 0) return;        if(target == 0){            result.add(new ArrayList<Integer>(item));        }        for(int i = start; i < candidates.length; i++){            //if(target < candidates[i]) return;            if(i > 0 && candidates[i] == candidates[i-1]) continue;            item.add(candidates[i]);            dfs(candidates, i, item, target-candidates[i], result);            item.remove(item.size()-1);        }    }

相关题目：8.5 Combinations Note: In Combinations, parameters in recursive call is i+1, not i.