目录
- 问题
- 示例
- 分析
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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] ]
示例Input: candidates = [2,3,5],
target = 8,
A solution set is: [ [2,2,2,2], [2,3,3], [3,5] ]
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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:
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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(n^2) O(n2) 。