我有这个数组,例如(大小是可变的):
x = ["10111", "10122", "10250", "10113"]
我需要找到最长的字符串,它是每个数组元素的子字符串(在本例中为“10”)(它不必是字符串的前缀)。我必须从所有字符串中删除它。此示例的输出为:
x=["111","222","250","113"] //common value = "10"
答案 0 :(得分:7)
此扩展名找到最常见的子串。请注意,"1"
也比"10"
更频繁地包含在每个字符串中。 (仅限C#):
public static class StringExtensions
{
public static IEnumerable<string> GetMostCommonSubstrings(this IList<string> strings)
{
if (strings == null)
throw new ArgumentNullException("strings");
if (!strings.Any() || strings.Any(s => string.IsNullOrEmpty(s)))
throw new ArgumentException("None string must be empty", "strings");
var allSubstrings = new List<List<string>>();
for (int i = 0; i < strings.Count; i++)
{
var substrings = new List<string>();
string str = strings[i];
for (int c = 0; c < str.Length - 1; c++)
{
for (int cc = 1; c + cc <= str.Length; cc++)
{
string substr = str.Substring(c, cc);
if (allSubstrings.Count < 1 || allSubstrings.Last().Contains(substr))
substrings.Add(substr);
}
}
allSubstrings.Add(substrings);
}
if (allSubstrings.Last().Any())
{
var mostCommon = allSubstrings.Last()
.GroupBy(str => str)
.OrderByDescending(g => g.Key.Length)
.ThenByDescending(g => g.Count())
.Select(g => g.Key);
return mostCommon;
}
return Enumerable.Empty<string>();
}
}
现在很容易:
string[] x = new[] { "10111", "10122", "10250", "10113" };
string mostCommonSubstring = x.GetMostCommonSubstrings().FirstOrDefault();
if (mostCommonSubstring != null)
{
for (int i = 0; i < x.Length; i++)
x[i] = x[i].Replace(mostCommonSubstring, "");
}
Console.Write(string.Join(", ", x));
输出:
111, 122, 250, 113
编辑:如果您只想在不考虑出现频率的情况下找到最长的公共子字符串,可以使用HashSet<string>
使用此优化方法(O(n)操作) :
public static string GetLongestCommonSubstring(this IList<string> strings)
{
if (strings == null)
throw new ArgumentNullException("strings");
if (!strings.Any() || strings.Any(s => string.IsNullOrEmpty(s)))
throw new ArgumentException("None string must be empty", "strings");
var commonSubstrings = new HashSet<string>(strings[0].GetSubstrings());
foreach (string str in strings.Skip(1))
{
commonSubstrings.IntersectWith(str.GetSubstrings());
if (commonSubstrings.Count == 0)
return null;
}
return commonSubstrings.OrderByDescending(s => s.Length).First();
}
public static IEnumerable<string> GetSubstrings(this string str)
{
if (string.IsNullOrEmpty(str))
throw new ArgumentException("str must not be null or empty", "str");
for (int c = 0; c < str.Length - 1; c++)
{
for (int cc = 1; c + cc <= str.Length; cc++)
{
yield return str.Substring(c, cc);
}
}
}
以这种方式使用它:
string[] x = new[] { "101133110", "101233210", "102533010", "101331310" };
string longestCommon = x.GetLongestCommonSubstring(); // "10"
答案 1 :(得分:1)
试试这个:(我想常见的字符串应该在开头):
string[] x = {"10111","10222","10250","10113"};
string common = x[0];
foreach(var i in x){
while(!i.StartsWith(common)){
common = common.Substring(0,common.Length-1);
if(common == "") break;
}
}
x = x.Select(a=>a.Substring(common.Length)).ToArray();
答案 2 :(得分:1)
查找长度为1的子字符串出现的最大次数。这是一个简单的O(n ^ 2)搜索。调用此最大出现次数K。
在您的示例中,这是“1”,“0”和K = 5.
现在你知道长度为2的所有子串都不能出现在K个输入字符串之外。此外,任何出现K次的子字符串必须由出现K次的长度为1的子字符串组成。在长度为1的子串中搜索存在K次的长度为2的子串,这也是一个简单的O(n ^ 2)搜索。
重复更长的长度,直到K输入中不再存在子串。