检查字符串是否是回文

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而不是某些逻辑。

我怎样才能做到这一点？

样品：

public bool getStatus(string myString) { } 

示例： myString =“ankYkna”，因此如果我们将其分成两个子字符串，则为：left-part =“ank”。 右边=“ank”（反转后）。

Answer 1

 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); } 

Answer 2

纯娱乐：

 return myString.SequenceEqual(myString.Reverse()); 

Answer 3

 int length = myString.Length; for (int i = 0; i < length / 2; i++) { if (myString[i] != myString[length - i - 1]) return false; } return true; 

Answer 4

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; 

Answer 5

  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--; } } 

Answer 6

//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; } 

Answer 7

使用Linq单行代码

 public static bool IsPalindrome(string str) { return str.SequenceEqual(str.Reverse()); } 

Answer 8

这种方式在外观和流程方面都非常简洁。

 Func IsPalindrome = s => s.Reverse().Equals(s); 

Answer 9

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(); } 

Answer 10

  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"); } 

Answer 11

 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; } 

Answer 12

如果您只需要检测回文，可以使用正则表达式进行检测，如此处所述。 可能不是最有效的方法，但……

Answer 13

这是非常重要的，没有内置的方法为你做这件事，你必须自己编写。 您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

Answer 14

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; } 

Answer 15

 public Boolean IsPalindrome(string value) { var one = value.ToList(); var two = one.Reverse().ToList(); return one.Equals(two); } 

Answer 16

 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。

Answer 17

 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(); } } 

Answer 18

 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; } 

Answer 19

 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; } 

Answer 20

这是一个绝对简单的方法，

1. 将单词作为输入接收到方法中。
2. 将临时变量分配给原始值。
3. 循环显示初始单词，并将最后一个字符添加到正在构建的反转中，直到初始单词不再有字符为止。
4. 现在使用您创建的备用来保存原始值以与构造的副本进行比较。

这是一个很好的方式，因为你不必投注和双打。 你可以使用ToString（）方法将它们传递给它们的字符串表示forms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。

Answer 21

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()); } 

Answer 22

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，（ 参见维基百科文章 ）我建议这个解决方案：

 public class Palindrome { static IList Allowed = new List { '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 characters = new List(); 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()); } } 

Answer 23

 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(); } 

Answer 24

 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(); 

Answer 25

 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; } 

Answer 26

从dotnetperls使用这种方式

  using System; class Program { /// 
 /// Determines whether the string is a palindrome. /// 
 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)); } } } 

Answer 27

这是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; } 

Answer 28

 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()); } } }