
目录
问题
给定一个数组 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
中的每个数字在每个组合中只能使用一次。
说明:
- 所有数字(包括目标数)都是正整数。
- 解集不能包含重复的组合。
输入: candidates = [10,1,2,7,6,1,5],
target = 8,
所求解集为: [ [1, 7], [1, 2, 5], [2, 6], [1, 1, 6] ]
输入: candidates = [2,5,2,1,2],
target = 5,
所求解集为: [ [1,2,2], [5] ]
Given a collection of candidate numbers (candidates) and a target number (target), find all unique combinations in candidates where the candidate numbers sums to target.
Each number in candidates may only be used once in the combination.
Note:
- All numbers (including target) will be positive integers.
- The solution set must not contain duplicate combinations.
Input: candidates = [10,1,2,7,6,1,5],
target = 8,
A solution set is: [ [1, 7], [1, 2, 5], [2, 6], [1, 1, 6] ]
Input: candidates = [2,5,2,1,2],
target = 5,
A solution set is: [ [1,2,2], [5] ]
示例
public class Program { public static void Main(string[] args) { var candidates = new int[] { 10, 1, 2, 7, 6, 1, 5 }; var target = 8; var res = CombinationSum2(candidates, target); ShowArray(res); Console.ReadKey(); } private static void ShowArray(IList<IList<int>> candidates) { foreach(var candi in candidates) { foreach(var num in candi) { Console.Write($"{num} "); } Console.WriteLine(); } Console.WriteLine(); } public static IList<IList<int>> CombinationSum2(int[] candidates, int target) { var res = new List<IList<int>>(); if(candidates.Length == 1) { if(candidates[0] == target) { res.Add(candidates); return res; } } var candi = new List<int>(); Array.Sort(candidates); Combination(candidates, -1, 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) { if(!res.Any(r => r.SequenceEqual(candi))) { res.Add(candi); } return; } for(var i = start + 1; i < candidates.Length; i++) { candi.Add(candidates[i]); Combination(candidates, i, target - candidates[i], candi.ToList(), ref res); candi.RemoveAt(candi.Count - 1); while(i < candidates.Length - 1 && candidates[i] == candidates[i + 1]) i++; } } }
以上给出1种算法实现,以下是这个案例的输出结果:
1 2 5 1 7 1 1 6 2 6
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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