LeetCode 216. Combination Sum III （组合的和之三）

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]]

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题目标签：Array, Backtracking

　　题目给了我们 k 和 n， 让我们找到 从1到9里的 k 个数字组合，没有重复，加起来之和等于 n。这道题目与前面两个版本的基本思路一样，利用 backtracking来做，就是有一些条件不一样而已。这题规定了只能从1 到 9 里选数字，之前两个版本的题目都是给的array；还有就是 这一题 规定了我们 数字的数量要等于 k。所以只要做一些条件的修改就可以了，具体看code。

Java Solution:

Runtime beats 45.85%

完成日期：09/06/2017

关键词：Array，Backtracking

关键点：加入数字 -> 递归 -> 去除最后一个数字 来测试所有的组合可能性

 class Solution
 {
     public List<List<Integer>> combinationSum3(int k, int n)
     {
         List<List<Integer>> list = new ArrayList<>();
         int len = 10; // use only numbers from 1 to 9
         backtrack(list, new ArrayList<>(), n, 1, len, k);

         return list;
     }

     public boolean backtrack(List<List<Integer>> list, List<Integer> tempList,
             int remain, int start, int len, int size)
     {
         if(remain < 0 || tempList.size() > size) // if remain less than 0 or number limit exceeds
             return false;                         // no need to continue
         else if(remain == 0 && tempList.size() == size) // only add tempList into list
         {                                               // when remain = 0 and number limit size matches
             list.add(new ArrayList<>(tempList));
             return false;
         }
         else
         {
             for(int i=start; i<len; i++)
             {
                 boolean flag;
                 tempList.add(i);
                 flag = backtrack(list, tempList, remain - i, i+1, len, size); // i + 1 because we cannot use same number more than once
                 tempList.remove(tempList.size() - 1);

                 if(!flag) // if find a tempList answer or fail, no need to continue that loop
                     break; // because 1~9 is sorted and unique, the rest numbers are greater, they'll fail
             }

             return true;
         }
     }
 }

参考资料：N/A

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