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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