
目录
问题
给定一个无重复元素的数组 candidates
和一个目标数 target
,找出 candidates
中所有可以使数字和为 target
的组合。
candidates
中的数字可以无限制重复被选取。
说明:
- 所有数字(包括
target
)都是正整数。 - 解集不能包含重复的组合。
输入: candidates = [2,3,6,7],
target = 7,
所求解集为: [ [7], [2,2,3] ]
输入: candidates = [2,3,5],
target = 8,
所求解集为: [ [2,2,2,2], [2,3,3], [3,5] ]
Given a set of candidate numbers (candidates) (without duplicates) and a target number (target), find all unique combinations in candidates where the candidate numbers sums to target.
The same repeated number may be chosen from candidates unlimited number of times.
Note:
- All numbers (including target) will be positive integers.
- The solution set must not contain duplicate combinations.
Input: candidates = [2,3,6,7],
target = 7,
A solution set is: [ [7], [2,2,3] ]
Input: candidates = [2,3,5],
target = 8,
A solution set is: [ [2,2,2,2], [2,3,3], [3,5] ]
示例
public class Program { public static void Main(string[] args) { var candidates = new int[] { 2, 3, 6, 7 }; var target = 7; var res = CombinationSum(candidates, target); ShowArray(res); Console.ReadKey(); } private static void ShowArray(List<IList<int>> candidates) { foreach(var candi in candidates) { foreach(var num in candi) { Console.Write($"{num} "); } Console.WriteLine(); } Console.WriteLine(); } public static List<IList<int>> CombinationSum(int[] candidates, int target) { var res = new List<IList<int>>(); var candi = new List<int>(); Combination(candidates, 0, target, candi, ref res); return res; } public static void Combination(int[] candidates, int start, int target, List<int> candi, ref List<IList<int>> res) { if(target < 0) return; if(target == 0) { res.Add(candi); return; } for(var i = start; i < candidates.Length; i++) { candi.Add(candidates[i]); Combination(candidates, i, target - candidates[i], candi.ToList(), ref res); candi.RemoveAt(candi.Count - 1); } } }
以上给出1种算法实现,以下是这个案例的输出结果:
2 2 3 7
分析
显而易见, 以上算法的时间复杂度为:O(n^2)O(n2) 。
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