Leetcode[39, 40, 216]Combination Sum

###Task1 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]

###Python

class Solution(object):
    def combinationSum(self, candidates, target):
        """
        :type candidates: List[int]
        :type target: int
        :rtype: List[List[int]]
        """
        ret = []
        temp = []
        self.helper(ret, temp, sorted(candidates), target)
        return ret
    
    def helper(self, ret, temp, candidates, target):
        if target == 0:
            ret.append(temp[:])
            return 
        for i, num in enumerate(candidates):
            if num > target:
                return
            temp.append(num)
            self.helper(ret, temp, candidates[i:], target - num)
            temp.pop()
        
        

###Java

public class Solution {
	public void combinator(int[] candidates, int target, ArrayList<List<Integer>> res, int pos, ArrayList<Integer> temp) {
		if (target == 0) {
			//cannot add temp directly !! its a object reference !!
			ArrayList<Integer> current = new ArrayList<Integer>(temp);
			res.add(current);
			return;
		}
		
		for (int i = pos; i < candidates.length; i++) {
			if (candidates[i] > target) {
				return;
			}
			
			temp.add(candidates[i]);
			combinator(candidates, target - candidates[i], res, i, temp);
			temp.remove(temp.size() - 1);
		}
	}
	
    public List<List<Integer>> combinationSum(int[] candidates, int target) {
        //edge cases
    	if (candidates == null || candidates.length == 0) {
    		return null;
    	}
    	
    	Arrays.sort(candidates);
    	ArrayList<List<Integer>> res = new ArrayList<List<Integer>>();
    	
    	ArrayList<Integer> temp = new ArrayList<Integer>();
    	
    	combinator(candidates, target, res, 0, temp);
    	
    	return res;      
    }
}

###Task2 Given a collection of candidate numbers (C) and a target number (T), find all unique combinations in C where the candidate numbers sums to T.

Each number in C may only be used once in the combination.

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 10,1,2,7,6,1,5 and target 8, A solution set is: [1, 7] [1, 2, 5] [2, 6] [1, 1, 6]

###Python ```pythonclass Solution(object): def combinationSum2(self, candidates, target): “”” :type candidates: List[int] :type target: int :rtype: List[List[int]] “”” ret = [] temp = [] self.helper(ret, temp, sorted(candidates), target) return ret def helper(self, ret, temp, candidates, target): if target == 0: ret.append(temp[:]) return for i, num in enumerate(candidates): if num > target: return if i != 0 and candidates[i] == candidates[i - 1]: continue temp.append(num) self.helper(ret, temp, candidates[i + 1:], target - num) temp.pop()

###Java
```java
public class Solution {
	public void combinationHelper(int[] candidates, List<List<Integer>> res, List<Integer> temp, int target, int pos) {
		if (target == 0) {
			res.add(new ArrayList<Integer>(temp));
			return;
		}
		for (int i = pos; i < candidates.length; i++) {
			if (candidates[i] > target) {
				break;
			}
			if (i != pos && candidates[i] == candidates[i - 1]) {
				continue;
			}
			temp.add(candidates[i]);
			combinationHelper(candidates, res, temp, target - candidates[i], i + 1);
			temp.remove(temp.size() - 1);
		}
		return;
	}

	 public List<List<Integer>> combinationSum2(int[] candidates, int target) {
	 	if (candidates == null || candidates.length == 0) {
	 		return null;
	 	}
	 	Arrays.sort(candidates);
	 	List<List<Integer>> res = new ArrayList<List<Integer>>();
	 	List<Integer> temp = new ArrayList<Integer>();

	 	combinationHelper(candidates, res, temp, target, 0);

	 	return res;
	 }
}

###Task3 Find all possible combinations of k numbers that add up to a number n, given that only numbers from 1 to 9 can be used and each combination should be a unique set of numbers.

Ensure that numbers within the set are sorted in ascending order.

Example 1:

Input: k = 3, n = 7

Output:

[[1,2,4]]

Example 2:

Input: k = 3, n = 9

Output:

[[1,2,6], [1,3,5], [2,3,4]]

###Python

class Solution(object):
    def combinationSum3(self, k, n):
        """
        :type k: int
        :type n: int
        :rtype: List[List[int]]
        """
        ret = []
        self.dfs(ret, k, n, [], 0)
        return ret
    def dfs(self, ret, k, target, temp, cur):
        if len(temp) == k and target == 0:
            ret.append(temp[:])
            return
        for i in range(cur + 1, 10):
            temp.append(i)
            self.dfs(ret, k, target - i, temp, i)
            temp.pop()
        

###Java

public class Solution {
	public void combinationHelper(List<List<Integer>> res, List<Integer> temp, int k, int n, int pos) {
		if (temp.size() == k && n == 0) {
			res.add(new ArrayList<Integer>(temp));
			return;
		}
		for (int i = pos; i <= 9; i++) {
			if (i > n) {
				break;
			}
			temp.add(i);
			combinationHelper(res, temp, k, n - i, i + 1);
			temp.remove(temp.size() - 1);
		}

		return;
	}

    public List<List<Integer>> combinationSum3(int k, int n) {
        List<List<Integer>> res = new ArrayList<List<Integer>>();
        List<Integer> temp = new ArrayList<Integer>();

        combinationHelper(res, temp, k, n, 1);

        return res;
    }
}