Combination Sum III

最新推荐文章于 2022-04-16 09:38:36 发布

sjphiChina

最新推荐文章于 2022-04-16 09:38:36 发布

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本文介绍了一种寻找所有可能的组合使得k个数加起来等于n的算法实现，使用回溯法并限定数字范围为1到9，确保每组组合唯一且不重复。通过具体的例子展示了算法的工作原理。

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挺简单的一道题，也出了3个错误，如何bug free啊

public class Solution {
    public List<List<Integer>> combinationSum3(int k, int n) {
        List<List<Integer>> result = new LinkedList<>();
        if (k == 0 || n < 1) {
            return result;
        }
        List<Integer> list = new LinkedList<>();
        //1 combinationSum3Helper(result, list, k, n, 0);
        combinationSum3Helper(result, list, k, n, 1);
        return result;
    }
    
    private void combinationSum3Helper(List<List<Integer>> result, List<Integer> list, int k, int n, int pos) {
        if (k == 0 && n == 0) {
            result.add(new LinkedList<>(list));
            return;
        } else if (n < 0 || k < 0) {
            return;
        }
        //2 for (int i = pos; i <= n; i++) {
        for (int i = pos; i <= 9; i++) {
            list.add(i);
            //3 ombinationSum3Helper(result, list, k - 1, n - i, i++);
            combinationSum3Helper(result, list, k - 1, n - i, i + 1);
            list.remove(list.size() - 1);
        }
        
    }
}

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: