39. Combination Sum

Given a set of candidate numbers (candidates) (without duplicates) and a target number (target), find all unique combinations in candidates where the candidate numbers sums to target.

The same repeated number may be chosen from candidates unlimited number of times.

Note:

• All numbers (including target) will be positive integers.
• The solution set must not contain duplicate combinations.

Example 1:

Input: candidates = [2,3,6,7], target = 7,
A solution set is:
[
  [7],
  [2,2,3]
]

Example 2:

Input: candidates = [2,3,5], target = 8,
A solution set is:
[
  [2,2,2,2],
  [2,3,3],
  [3,5]
]

backtracking

每次递归到某个元素的时候，tmp中加入当前元素，target减去当前元素，当target=0的时候，把tmp加入res中

剪枝：排序+递归过程中判断当前元素是否大于target (大于就break)

s = target / min(nums[i]), T = C(s, 1) + C(s, 2) + ... + C(s, s) = 2^s  -> O(2^s)

time: O(2^s), space: O( target / min(nums[i]) )

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

or

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