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.

Note:
All numbers will be positive integers.
The solution set must not contain duplicate combinations.
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]]

难度:medium

题目:找出所有可能的K个数之和为n的组合,只可以使用1到9之间的数。
注意:所有数都为正,给出的答案中不包含重复的数。

思路:递归

Runtime: 1 ms, faster than 66.65% of Java online submissions for Combination Sum III.
Memory Usage: 35 MB, less than 54.02% of Java online submissions for Combination Sum III.

class Solution {
    public List<List<Integer>> combinationSum3(int k, int n) {
        List<List<Integer>> result = new ArrayList<>();
        if (n > 45 || n < 1 || k < 1 || k > 9) {
            return result;
        }
        
        combination(1, k, 0, n, new Stack<>(), result);
        
        return result;
    }
    
    private void combination(int begin, int k, int sum, int n,
                             Stack<Integer> stack, List<List<Integer>> result) {
        if (k < 0 || sum > n) {
            return;
        }
        if (0 == k && sum == n) {
            result.add(new ArrayList<>(stack));
            return;
        }
        
        for (int i = begin; i <= 9 - k + 1; i++) {
            stack.push(i);
            combination(i + 1, k - 1, sum + i, n, stack, result);
            stack.pop();
        }
    }
}

linm
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