目标和问题是找到一个子集的问题,使得元素的总和等于给定的数字。回溯方法在最坏的情况下生成所有排列,但一般来说,在解决子集和问题时比递归方法表现更好。
给定n个正整数的子集A和一个值和,求给定集合的任何子集是否存在,其元素之和等于给定的sum值
假设我们有一个数组 [1,2,3] 输出将是“1,1,1,1”,“1,1,2”,“2,2”,“13” 从输出“31”, ”211”、”121”可以丢弃
using System; using System.Collections.Generic; using System.Text; using System.Linq; namespace ConsoleApplication{ public class BackTracking{ public void Combinationsums(int[] array, int target){ List<int> currentList = new List<int>(); List<List<int>> results = new List<List<int>>(); int sum = 0; int index = 0; CombinationSum(array, target, currentList, results, sum, index); foreach (var item in results){ StringBuilder s = new StringBuilder(); foreach (var item1 in item){ s.Append(item1.ToString()); } Console.WriteLine(s); s = null; } } private void CombinationSum(int[] array, int target, List<int> currentList, List<List<int>> results, int sum, int index){ if (sum > target){ return; } else if (sum == target){ if (!results.Contains(currentList)){ List<int> newList = new List<int>(); newList.AddRange(currentList); results.Add(newList); return; } } else{ for (int i = 0; i < array.Length; i++){ currentList.Add(array[i]); CombinationSum(array, target, currentList, results, sum + array[i], i); currentList.Remove(array[i]); } } } } class Program{ static void Main(string[] args){ BackTracking b = new BackTracking(); int[] arrs = { 1, 2, 3 }; b.Combinationsums(arrs, 4); } } }输出结果
1111 112 13 22