216. Combination Sum III

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

因为k是不定的, 所以无法用LOOP（循环）. 回溯法是此类题的常用解法. 注意要新建一个list 放入结果中, 否则放入的reference 会指向原来的不断变化的list,result.add(new ArrayList(cur));

代码

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){        //难点在于如何取得所有不重复的组合数        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);        }    }}

结果及分析

递归这种分析时间复杂度的就是看最后遍历的次数，回溯法难就难在不好理解遍历的顺序。此次遍历的顺序是：[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

思路还是一样的，利用回溯法和dfs，code更简单了一点。

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);        }    }}