算法笔记 40. Combination Sum II via Java

class Solution {
    // using backtracking to build solution set;
    // traverse through the candidates list and try using each element as start point for building solution

    public List<List<Integer>> combinationSum2(int[] candidates, int target) {
        List<List<Integer>> res = new ArrayList<>(); // solution set, in which are 
        List<Integer> build = new ArrayList<>();    // partial solution
        Arrays.sort(candidates);
        helper(candidates, 0, build, res, target);
        return res;
    }
    private void helper(int[] candidates, int idx, List<Integer> build, 
                        List<List<Integer>> res, int target){
        if(target==0){
            List<Integer> toAdd = new ArrayList<>(build);
            res.add(toAdd);
            return;
        }
        for(int i = idx; i<candidates.length; i++){
            if(candidates[i]<=target){
                build.add(candidates[i]);
                helper(candidates, i+1, build, res, target-candidates[i]);
                build.remove(build.size()-1);
            }
            while(i<candidates.length-1 && candidates[i]==candidates[i+1]){
                i++;
            }
        }
    }
}

时间复杂度分析：排序O(nlgn) + n层递归，每个结点内最坏进行一个O(n)的遍历，递归树的形状不balance，不是很好分析，近似为O(2^n) 参考分析:https://www.1point3acres.com/bbs/thread-117602-1-1.html

空间复杂度：递归树上下一条路径中所有节点的空间和，每层结点中并没有用到额外空间，所以这部分时O(height)=O(n)，保存答案空间复杂度为O(n)（不确定）总体为O(n)