下面的代码根据1,2,3 = 3,2,1 = 2,3,1的逻辑返回所有不同的组合,因此它只返回该组数字的1个实例.
但是,我想更改该逻辑,以便它返回所有数字集的所有实例.
为了实现这一目标,我需要对“GetPowerSet”下面的LINQ查询做些什么?
public void GetPowersets()
{
List<int> ints = new List<int>()
{
1,2,2,3,3
};
var results = GetPowerSet(ints);
List<String> combinations = new List<String>();
foreach (var result in results)
{
StringBuilder sb = new StringBuilder();
foreach (var intValue in result.OrderBy(x => x))
{
sb.Append(intValue + ",");
}
combinations.Add(sb.ToString());
}
string c1 = string.Join("|", combinations.ToArray()).Replace(",|", "|");
//c1 = "|1|2|1,2|2|1,2|2,2|1,2,2|3|1,3|2,3|1,2,3|2,3|1,2,3|2,2,3|1,2,2,3|3|1,3|2,3|1,2,3|2,3|1,2,3|2,2,3|1,2,2,3|3,3|1,3,3|2,3,3|1,2,3,3|2,3,3|1,2,3,3|2,2,3,3|1,2,2,3,3,"
}
public IEnumerable<IEnumerable<int>> GetPowerSet(List<int> list)
{
return from m in Enumerable.Range(0, 1 << list.Count)
select
from i in Enumerable.Range(0, list.Count)
where (m & (1 << i)) != 0
select list[i];
}
这是我想要达到的最终结果:(没有重复的组合行:duplicate = 3,2,1和3,2,1是相同的东西.但是1,2,3和3,2,1不是同样的事情,两者都应该在最终结果)
1
2
3
1,2
1,3
2,1
2,3
2,2
3,1
3,2
3,3
1,2,3
1,2,2
1,3,2
1,3,3
2,1,3
2,1,2
2,3,1
2,3,2
2,3,3
2,2,1
2,2,3
3,1,2
3,1,3
3,2,1
3,2,2
3,2,3
3,3,1
3,3,2
1,2,3,2
1,2,3,3
1,2,2,3
1,3,2,2
1,3,2,3
1,3,3,2
2,1,3,2
2,1,3,3
2,1,2,3
2,3,1,2
2,3,1,3
2,3,2,1
2,3,2,3
2,3,3,1
2,3,3,2
2,2,1,3
2,2,3,1
2,2,3,3
3,1,2,2
3,1,2,3
3,1,3,2
3,2,1,2
3,2,1,3
3,2,2,1
3,2,2,3
3,2,3,1
3,2,3,2
3,3,1,2
3,3,2,1
3,3,2,2
1,2,3,2,3
1,2,3,3,2
1,2,2,3,3
1,3,2,2,3
1,3,2,3,2
1,3,3,2,2
2,1,3,2,3
2,1,3,3,2
2,1,2,3,3
2,3,1,2,3
2,3,1,3,2
2,3,2,1,3
2,3,2,3,1
2,3,3,1,2
2,3,3,2,1
2,2,1,3,3
2,2,3,1,3
2,2,3,3,1
3,1,2,2,3
3,1,2,3,2
3,1,3,2,2
3,2,1,2,3
3,2,1,3,2
3,2,2,1,3
3,2,2,3,1
3,2,3,1,2
3,2,3,2,1
3,3,1,2,2
3,3,2,1,2
3,3,2,2,1
执行此操作的“foreach”方法,一旦数字集太大(我预期LINQ不应该出现此问题),往往会导致“Out of Memory Exception”.这是我想要的,返回我想要的结果集.但它很慢并且存在性能问题.我也愿意接受如何改善它的建议.
public List<List<int>> GetAllCombinationsOfAllSizes(List<int> ints)
{
List<List<int>> returnResult = new List<List<int>>();
var distinctInts = ints.Distinct().ToList();
for (int j = 0; j < distinctInts.Count(); j++)
{
var number = distinctInts[j];
var newList = new List<int>();
newList.Add(number);
returnResult.Add(newList);
var listMinusOneObject = ints.Select(x => x).ToList();
listMinusOneObject.Remove(listMinusOneObject.Where(x => x == number).First());
if (listMinusOneObject.Count() > 0)
{
_GetAllCombinationsOfAllSizes(listMinusOneObject, newList, ref returnResult);
}
}
return returnResult;
}
public void _GetAllCombinationsOfAllSizes(List<int> ints, List<int> growingList, ref List<List<int>> returnResult)
{
var distinctInts = ints.Distinct().ToList();
for (int j = 0; j < distinctInts.Count(); j++)
{
var number = distinctInts[j];
var newList = growingList.ToList();
newList.Add(number);
returnResult.Add(newList);
var listMinusOneObject = ints.Select(x => x).ToList();
listMinusOneObject.Remove(listMinusOneObject.Where(x => x == number).First());
if (listMinusOneObject.Count() > 0)
{
_GetAllCombinationsOfAllSizes(listMinusOneObject, newList, ref returnResult);
}
}
}
我正在寻找的答案是:如何实现我想要的结果集,但是使用LINQ和C#来实现它,其方式比我发布的当前“foreach”方式更快,更高效?
最佳答案 NEW UPDATE(删除旧代码,性能优于OP代码,产生输出)
static IEnumerable<int[]> EnumeratePermutations2(int[] ints)
{
Dictionary<int, int> intCounts = ints.GroupBy(n => n)
.ToDictionary(g => g.Key, g => g.Count());
int[] distincts = intCounts.Keys.ToArray();
foreach (int[] permutation in EnumeratePermutations2(new int[0], intCounts, distincts))
yield return permutation;
}
static IEnumerable<int[]> EnumeratePermutations2(int[] prefix, Dictionary<int, int> intCounts, int[] distincts)
{
foreach (int n in distincts)
{
int[] newPrefix = new int[prefix.Length + 1];
Array.Copy(prefix, newPrefix, prefix.Length);
newPrefix[prefix.Length] = n;
yield return newPrefix;
intCounts[n]--;
int[] newDistincts = intCounts[n] > 0
? distincts
: distincts.Where(x => x != n).ToArray();
foreach (int[] permutation in EnumeratePermutations2(newPrefix, intCounts, newDistincts))
yield return permutation;
intCounts[n]++;
}
}