例1: 输入 k = 3, n = 7; 输出 [[1,2,4]]

例2: 输入 k = 3, n = 9; 输出 [[1,2,6], [1,3,5], [2,3,4]]

using System;using System.Collections.Generic;public class Solution{    /// <summary>    /// 从1到9中，选择k个不同的数字，使其和等于n    /// </summary>    /// <param name="k">选择的数字个数</param>    /// <param name="n">数字的和</param>    /// <returns>复合条件的数字集合</returns>    public IList<IList<int>> CombinationSum3(int k, int n)    {        int[] allNums = new int[10];        int[] tagNums = new int[10];        List<string> results = new List<string>();        Solution p = new Solution();        for (int i = 1; i < 10; i++)        {            allNums[i - 1] = i;        }        //这里备选数字集合为1-9这9个数        p.Combine(allNums, 9, k, tagNums, k, results);        IList<IList<int>> correctList = new List<IList<int>>();        foreach (var res in results)        {            int sum = 0;            List<int> correct = new List<int>();            //计算各串数字的和            for (int j = 0; j < res.Length; j++)            {                sum += Convert.ToInt16(res[j].ToString());            }            if (sum == n)            {                for (int m = 0; m <= res.Length - 1; m++)                {                    correct.Add(Convert.ToInt16(res[m].ToString()));                }                correctList.Add(correct);            }            //清空sum的值            sum = 0;        }        return correctList;    }    /// <summary>    /// 输出给定数字范围内的所有组合（非排列）结果    /// </summary>    /// <param name="allNums">备选数字集合</param>    /// <param name="needCount">在备选数字集合中，选择 needCount 个中数字来进行再次选择</param>    /// <param name="childCount">取 childCount 个数字进行组合</param>    /// <param name="tagNums">存放每次选中的 childCount 个数字在备选数字集合中的位置</param>    /// <param name="backupChildCount">存放 childCount 的值</param>    /// <param name="results">所有组合（非排列）结果</param>    public void Combine(int[] allNums, int needCount, int childCount, int[] tagNums, int backupChildCount, List<string> results)    {        for (int i = needCount; i >= childCount; i--)        {            //选择第一个数字，记录它在 allNums 数组中的位置            tagNums[childCount - 1] = i - 1;            //如果选中的数字个数未达到 childCount 个，则继续选择下一个            if (childCount > 1)            {                Combine(allNums, i - 1, childCount - 1, tagNums, backupChildCount, results);            }            else            {                string res = "";                for (int j = 0; j <= backupChildCount - 1; j++)                {                    res += allNums[tagNums[j]].ToString();                }                results.Add(res);            }        }    }}