Question

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.
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]]

思路

因為k是不定的, 所以無法用LOOP（迴圈）. 回溯法是此類題的常用解法. 注意要新建一個list 放入結果中, 否則放入的reference 會指向原來的不斷變化的list,result.add(new ArrayList(cur));

程式碼

public class Solution {
    public List<List<Integer>> combinationSum3(int k, int n) {
        List<List<Integer
 
>> result = new ArrayList<List<Integer>>();
        List<Integer> list = new ArrayList<Integer>();
        if(k < 1 || k > 9 || n < 1 || n > 45)
            return result;
        generate(result,list,1,k,n);
        return result;
    }
    public void generate(List
 
<List<Integer>> result,List<Integer> list,int start,int k,int n){
        //難點在於如何取得所有不重複的組合數
        if(list.size() == k && n == 0){
            result.add(new ArrayList(list));
            return;
        }
        else if(list.size() >= k || n <= 0)
            return;
        for(int i = start;i <= 9;i++){
            list.add(i);
            generate(result,list,i + 1,k,n - i);
            list.remove(list.size() - 1);
        }
    }
}

結果及分析

遞迴這種分析時間複雜度的就是看最後遍歷的次數，回溯法難就難在不好理解遍歷的順序。此次遍歷的順序是：[1,2,3]->[1,2,4]->[1,2,5]->[1,2,6]->[1,2,7]->[1,2,8]->[1,2,9]->[1,3,4]->[1,3,5]…….

二刷

思路還是一樣的，利用回溯法和dfs，code更簡單了一點。

CODE

public class Solution {
    public List<List<Integer>> combinationSum3(int k, int n) {
        List<List<Integer>> res = new ArrayList<>();
        if(k == 0 || n == 0)
            return res;
        List<Integer> temp = new ArrayList<>();
        helper(res,temp,k,n,1);
        return res;
    }
    public void helper(List<List<Integer>> res,List<Integer> temp,int k,int n,int num){
        if(temp.size() == k && n == 0){
            res.add(new ArrayList<>(temp));
            return;
        }
        else if(temp.size() > k || n < 0)
            return;
        for(int i = num;i <= 9;i++){
            if(i > n)
                return;
            temp.add(i);
            n -= i;
            helper(res,temp,k,n,i + 1);
            n += i;
            temp.remove(temp.size() - 1);
        }
    }
}