我有一个字符串作为输入，必须打破两个子串中的字符串。如果左子串等于右子串，则执行一些逻辑。
我怎样才能做到这一点？
样品：
public bool getStatus(string myString)
{
}
例如：myString = "ankYkna"，所以如果我们把它分成两个子串，那就是：left-part = "ank"，right-part = "ank"（逆转之后）。
public static bool getStatus(string myString)
{
string first = myString.Substring(0, myString.Length / 2);
char[] arr = myString.ToCharArray();
Array.Reverse(arr);
string temp = new string(arr);
string second = temp.Substring(0, temp.Length / 2);
return first.Equals(second);
}
字符串扩展方法，使用方便：
public static bool IsPalindrome(this string str)
{
str = new Regex("[^a-zA-Z]").Replace(str, "").ToLower();
return !str.Where((t, i) => t != str[str.Length - i - 1]).Any();
}
private void CheckIfPalindrome(string str)
{
//place string in array of chars
char[] array = str.ToCharArray();
int length = array.Length -1 ;
Boolean palindrome =true;
for (int i = 0; i <= length; i++)//go through the array
{
if (array[i] != array[length])//compare if the char in the same positions are the same eg "tattarrattat" will compare array[0]=t with array[11] =t if are not the same stop the for loop
{
MessageBox.Show("not");
palindrome = false;
break;
}
else //if they are the same make length smaller by one and do the same
{
length--;
}
}
if (palindrome) MessageBox.Show("Palindrome");
}
如果您只是需要检测回文，您可以使用正则表达式进行检查，如here所述。可能不是最有效的方法，但......
这是非常重要的，没有内置的方法可以为你做到这一点，你必须自己编写。您将需要考虑要检查的规则，例如您隐含地声明您接受了一个字符串的反转。而且，你错过了中间字符，这只是奇数长度？
所以你会有类似的东西：
if(myString.length % 2 = 0)
{
//even
string a = myString.substring(0, myString.length / 2);
string b = myString.substring(myString.length / 2 + 1, myString.lenght/2);
if(a == b)
return true;
//Rule 1: reverse
if(a == b.reverse()) //can't remember if this is a method, if not you'll have to write that too
return true;
等，也为奇数字符串做任何你想做的事
这个C＃方法将检查偶数和奇数长度的回文串（递归方法）：
public static bool IsPalindromeResursive(int rightIndex, int leftIndex, char[] inputString)
{
if (rightIndex == leftIndex || rightIndex < leftIndex)
return true;
if (inputString[rightIndex] == inputString[leftIndex])
return IsPalindromeResursive(--rightIndex, ++leftIndex, inputString);
else
return false;
}
public Boolean IsPalindrome(string value)
{
var one = value.ToList<char>();
var two = one.Reverse<char>().ToList();
return one.Equals(two);
}
protected bool CheckIfPalindrome(string text)
{
if (text != null)
{
string strToUpper = Text.ToUpper();
char[] toReverse = strToUpper.ToCharArray();
Array.Reverse(toReverse );
String strReverse = new String(toReverse);
if (strToUpper == toReverse)
{
return true;
}
else
{
return false;
}
}
else
{
return false;
}
}
用这个最简单的方式。
class Program
{
static void Main(string[] args)
{
string s, revs = "";
Console.WriteLine(" Enter string");
s = Console.ReadLine();
for (int i = s.Length - 1; i >= 0; i--) //String Reverse
{
Console.WriteLine(i);
revs += s[i].ToString();
}
if (revs == s) // Checking whether string is palindrome or not
{
Console.WriteLine("String is Palindrome");
}
else
{
Console.WriteLine("String is not Palindrome");
}
Console.ReadKey();
}
}
public bool IsPalindroom(string input)
{
input = input.ToLower();
var loops = input.Length / 2;
var higherBoundIdx = input.Length - 1;
for (var lowerBoundIdx = 0; lowerBoundIdx < loops; lowerBoundIdx++, higherBoundIdx--)
{
if (input[lowerBoundIdx] != input[higherBoundIdx])
return false;
}
return true;
}
这是一个绝对简单的方法，
这是一个很好的方式，因为你不必抛出整数和双精度。你可以使用ToString（）方法将它们传递给它们的字符串表示形式的方法。
public static bool IsPalindrome(string word)
{
string spare = word;
string reversal = null;
while (word.Length > 0)
{
reversal = string.Concat(reversal, word.LastOrDefault());
word = word.Remove(word.Length - 1);
}
return spare.Equals(reversal);
}
所以从你的main方法来看，对于偶数和奇数长度的字符串，你只需将整个字符串传递给方法。
纯娱乐：
return myString.SequenceEqual(myString.Reverse());
在所有解决方案中，也可以尝试以下方法：
public static bool IsPalindrome(string s)
{
return s == new string(s.Reverse().ToArray());
}
由于回文也包括数字，单词，句子以及这些的任何组合，并且应该忽略标点符号和案例，（See Wikipedia Article）我建议这个解决方案：
public class Palindrome
{
static IList<int> Allowed = new List<int> {
'a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'h',
'i', 'j', 'k', 'l', 'm', 'n', 'o', 'p', 'q',
'r', 's', 't', 'u', 'v', 'w', 'x', 'y', 'z',
'1', '2', '3', '4', '5', '6', '7', '8', '9',
'0'
};
private static int[] GetJustAllowed(string text)
{
List<int> characters = new List<int>();
foreach (var c in text)
characters.Add(c | 0x20);
return characters.Where(c => Allowed.Contains(c)).ToArray();
}
public static bool IsPalindrome(string text)
{
if(text == null || text.Length == 1)
return true;
int[] chars = GetJustAllowed(text);
var length = chars.Length;
while (length > 0)
if (chars[chars.Length - length] != chars[--length])
return false;
return true;
}
public static bool IsPalindrome(int number)
{
return IsPalindrome(number.ToString());
}
public static bool IsPalindrome(double number)
{
return IsPalindrome(number.ToString());
}
public static bool IsPalindrome(decimal number)
{
return IsPalindrome(number.ToString());
}
}
static void Main(string[] args)
{
string str, rev="";
Console.Write("Enter string");
str = Console.ReadLine();
for (int i = str.Length - 1; i >= 0; i--)
{
rev = rev + str[i];
}
if (rev == str)
Console.Write("Entered string is pallindrome");
else
Console.Write("Entered string is not pallindrome");
Console.ReadKey();
}
string test = "Malayalam";
char[] palindrome = test.ToCharArray();
char[] reversestring = new char[palindrome.Count()];
for (int i = palindrome.Count() - 1; i >= 0; i--)
{
reversestring[palindrome.Count() - 1 - i] = palindrome[i];
}
string materializedString = new string(reversestring);
if (materializedString.ToLower() == test.ToLower())
{
Console.Write("Palindrome!");
}
else
{
Console.Write("Not a Palindrome!");
}
Console.Read();
public static bool palindrome(string t)
{
int i = t.Length;
for (int j = 0; j < i / 2; j++)
{
if (t[j] == t[i - j-1])
{
continue;
}
else
{
return false;
break;
}
}
return true;
}
从dotnetperls使用这种方式
using System;
class Program
{
/// <summary>
/// Determines whether the string is a palindrome.
/// </summary>
public static bool IsPalindrome(string value)
{
int min = 0;
int max = value.Length - 1;
while (true)
{
if (min > max)
{
return true;
}
char a = value[min];
char b = value[max];
// Scan forward for a while invalid.
while (!char.IsLetterOrDigit(a))
{
min++;
if (min > max)
{
return true;
}
a = value[min];
}
// Scan backward for b while invalid.
while (!char.IsLetterOrDigit(b))
{
max--;
if (min > max)
{
return true;
}
b = value[max];
}
if (char.ToLower(a) != char.ToLower(b))
{
return false;
}
min++;
max--;
}
}
static void Main()
{
string[] array =
{
"A man, a plan, a canal: Panama.",
"A Toyota. Race fast, safe car. A Toyota.",
"Cigar? Toss it in a can. It is so tragic.",
"Dammit, I'm mad!",
"Delia saw I was ailed.",
"Desserts, I stressed!",
"Draw, O coward!",
"Lepers repel.",
"Live not on evil.",
"Lonely Tylenol.",
"Murder for a jar of red rum.",
"Never odd or even.",
"No lemon, no melon.",
"Senile felines.",
"So many dynamos!",
"Step on no pets.",
"Was it a car or a cat I saw?",
"Dot Net Perls is not a palindrome.",
"Why are you reading this?",
"This article is not useful.",
"...",
"...Test"
};
foreach (string value in array)
{
Console.WriteLine("{0} = {1}", value, IsPalindrome(value));
}
}
}
这是一种检查回文的简短而有效的方法。
bool checkPalindrome(string inputString) {
int length = inputString.Length;
for(int i = 0; i < length/2; i++){
if(inputString[i] != inputString[length-1-i]){
return false;
}
}
return true;
}
public bool Solution(string content)
{
int length = content.Length;
int half = length/2;
int isOddLength = length%2;
// Counter for checking the string from the middle
int j = (isOddLength==0) ? half:half+1;
for(int i=half-1;i>=0;i--)
{
if(content[i] != content[j])
{
return false;
}
j++;
}
return true;
}
public bool MojTestPalindrome (string word)
{
bool yes = false;
char[]test1 = word.ToArray();
char[] test2 = test1.Reverse().ToArray();
for (int i=0; i< test2.Length; i++)
{
if (test1[i] != test2[test2.Length - 1 - i])
{
yes = false;
break;
}
else {
yes = true;
}
}
if (yes == true)
{
return true;
}
else
return false;
}
在C＃中：
public bool EhPalindromo(string text)
{
var reverseText = string.Join("", text.ToLower().Reverse());
return reverseText == text;
}
int length = myString.Length;
for (int i = 0; i < length / 2; i++)
{
if (myString[i] != myString[length - i - 1])
return false;
}
return true;
using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Threading.Tasks;
class palindrome
{
static void Main(string[] args)
{
Console.Write("Enter a number:");
string panstring = Console.ReadLine();
Palindrome(panstring);
Console.ReadKey();
}
static int index = 0;
public static void Palindrome(string strexcluding)
{
try
{
string reversecounter = string.Empty;
for (int i = strexcluding.Length - 1; i >= 0; i--)
{
if (strexcluding[i].ToString() != null)
reversecounter += strexcluding[i].ToString();
}
if (reversecounter == strexcluding)
{
Console.WriteLine("Palindrome Number: " + strexcluding);
}
else
{
Sum(strexcluding);
}
}
catch(Exception ex)
{
Console.WriteLine(ex.ToString());
}
}
public static void Sum(string stringnumber)
{
try
{
index++;
string number1 = stringnumber;
string number2 = stringnumber;
string[] array = new string[number1.Length];
string obtained = string.Empty;
string sreverse = null;
Console.WriteLine(index + ".step : " + number1 + "+" + number2);
for (int i = 0; i < number1.Length; i++)
{
int temp1 = Convert.ToInt32(number1[number1.Length - i - 1].ToString());
int temp2 = Convert.ToInt32(number2[number2.Length - i - 1].ToString());
if (temp1 + temp2 >= 10)
{
if (number1.Length - 1 == number1.Length - 1 - i)
{
array[i] = ((temp1 + temp2) - 10).ToString();
obtained = "one";
}
else if (number1.Length - 1 == i)
{
if (obtained == "one")
{
array[i] = (temp1 + temp2 + 1).ToString();
}
else
{
array[i] = (temp1 + temp2).ToString();
}
}
else
{
if (obtained == "one")
{
array[i] = ((temp1 + temp2 + 1) - 10).ToString();
}
else
{
array[i] = ((temp1 + temp2) - 10).ToString();
obtained = "one";
}
}
}
else
{
if (obtained == "one")
array[i] = (temp1 + temp2 + 1).ToString();
else
array[i] = (temp1 + temp2).ToString();
obtained = "Zero";
}
}
for (int i = array.Length - 1; i >= 0; i--)
{
if (array[i] != null)
sreverse += array[i].ToString();
}
Palindrome(sreverse);
}
catch(Exception ex)
{
Console.WriteLine(ex.ToString());
}
}
}
使用LINQ并且远离最佳解决方案
var original = "ankYkna";
var reversed = new string(original.Reverse().ToArray());
var palindrom = original == reversed;
使用Linq的一行代码
public static bool IsPalindrome(string str)
{
return str.SequenceEqual(str.Reverse());
}
public static bool IsPalindrome(string value)
{
int i = 0;
int j = value.Length - 1;
while (true)
{
if (i > j)
{
return true;
}
char a = value[i];
char b = value[j];
if (char.ToLower(a) != char.ToLower(b))
{
return false;
}
i++;
j--;
}
}
//这个c＃方法将检查偶数和奇数长度的回文串
public static bool IsPalenDrome(string palendromeString)
{
bool isPalenDrome = false;
try
{
int halfLength = palendromeString.Length / 2;
string leftHalfString = palendromeString.Substring(0,halfLength);
char[] reversedArray = palendromeString.ToCharArray();
Array.Reverse(reversedArray);
string reversedString = new string(reversedArray);
string rightHalfStringReversed = reversedString.Substring(0, halfLength);
isPalenDrome = leftHalfString == rightHalfStringReversed ? true : false;
}
catch (Exception ex)
{
throw ex;
}
return isPalenDrome;
}
这种方式在外观和流程方面都非常简洁。
Func<string, bool> IsPalindrome = s => s.Reverse().Equals(s);
public static bool IsPalindrome(string word)
{
//first reverse the string
string reversedString = new string(word.Reverse().ToArray());
return string.Compare(word, reversedString) == 0 ? true : false;
}