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] 

DFS: 

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

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] 

基本思路和permutation2一样，需要一个boolean array来帮助检测重复情况。

public class Solution {
    //Time: O(n^n)  
    public ArrayList<ArrayList<Integer>> combinationSum2(int[] num, int target) {
        ArrayList<ArrayList<Integer>> res = new ArrayList<ArrayList<Integer>>();
        ArrayList<Integer> tmp = new ArrayList<Integer>();
        Arrays.sort(num);
        boolean[] visit = new boolean[num.length];
        combination(res, tmp, num, visit, target, 0);
        return res;
    }
    
    private void combination(ArrayList<ArrayList<Integer>> res, ArrayList<Integer> tmp,
                             int[] num, boolean[] visit, int target, int index) {
        if (target < 0) {
            return;
        }  
        
        if (target == 0) {
            res.add(new ArrayList<Integer>(tmp));
            return;
        }
        
        for (int i = index; i < num.length; i++) {
            if (i != 0 && num[i] == num[i - 1] && !visit[i - 1]) {
                continue;
            }
            
            tmp.add(num[i]);
            visit[i] = true;
            combination(res, tmp, num, visit, target - num[i], i + 1);
            tmp.remove(tmp.size() - 1);
            visit[i] = false;
        }
    }
}