字典序上的第一个回文字符串
重新排列给定字符串的字符以形成按字典顺序排列的第一个回文字符串。如果不存在这样的字符串，则显示消息“没有回文字符串”。
例子：
Input : malayalam
Output : aalmymlaa
Input : apple
Output : no palindromic string
简单的方法：
1. 按字母（升序）顺序对字符串字符进行排序。
2. 一个是按字典顺序查找给定字符串的下一个排列。
3. 第一个回文排列就是答案。
有效方法：回文字符串的属性：
1. 如果字符串的长度是偶数，那么字符串中每个字符出现的频率一定是偶数。
2. 如果长度为奇数，则应该有一个字符的频率为奇数，所有其他字符的频率必须为偶数，并且字符串中间必须至少出现一次奇数字符。
算法
1. 存储给定字符串中每个字符的频率
2. 使用上述回文字符串的属性，检查是否可以形成回文字符串。
3、如果不能形成回文字符串，返回“No Palindromic String”。
4. 否则我们创建三个字符串，然后返回front_str + odd_str + rear_str。
- odd_str ：如果没有奇数频率的字符，则为空。否则它包含所有出现的奇数字符。
- front_str ：按升序包含字符串中所有偶数出现的字符的一半。
- back_str包含字符串中所有偶数出现的字符的一半，与 front_str 的顺序相反。
下面是上述步骤的实现。
C++
// C++ program to find first palindromic permutation // of given string #includeusing namespace std; const char MAX_CHAR = 26; // Function to count frequency of each char in the // string. freq[0] for 'a',...., freq[25] for 'z' void countFreq(string str, int freq[], int len) { for (int i=0; i Java
// Java program to find first palindromic permutation // of given string class GFG { static char MAX_CHAR = 26; // Function to count frequency of each char in the // string. freq[0] for 'a',...., freq[25] for 'z' static void countFreq(String str, int freq[], int len) { for (int i = 0; i < len; i++) { freq[str.charAt(i) - 'a']++; } } // Cases to check whether a palindr0mic // string can be formed or not static boolean canMakePalindrome(int freq[], int len) { // count_odd to count no of // chars with odd frequency int count_odd = 0; for (int i = 0; i < MAX_CHAR; i++) { if (freq[i] % 2 != 0) { count_odd++; } } // For even length string // no odd freq character if (len % 2 == 0) { if (count_odd > 0) { return false; } else { return true; } } // For odd length string // one odd freq character if (count_odd != 1) { return false; } return true; } // Function to find odd freq char and // reducing its freq by 1returns "" if odd freq // char is not present static String findOddAndRemoveItsFreq(int freq[]) { String odd_str = ""; for (int i = 0; i < MAX_CHAR; i++) { if (freq[i] % 2 != 0) { freq[i]--; odd_str = odd_str + (char) (i + 'a'); return odd_str; } } return odd_str; } // To find lexicographically first palindromic // string. static String findPalindromicString(String str) { int len = str.length(); int freq[] = new int[MAX_CHAR]; countFreq(str, freq, len); if (!canMakePalindrome(freq, len)) { return "No Palindromic String"; } // Assigning odd freq character if present // else empty string. String odd_str = findOddAndRemoveItsFreq(freq); String front_str = "", rear_str = " "; // Traverse characters in increasing order for (int i = 0; i < MAX_CHAR; i++) { String temp = ""; if (freq[i] != 0) { char ch = (char) (i + 'a'); // Divide all occurrences into two // halves. Note that odd character // is removed by findOddAndRemoveItsFreq() for (int j = 1; j <= freq[i] / 2; j++) { temp = temp + ch; } // creating front string front_str = front_str + temp; // creating rear string rear_str = temp + rear_str; } } // Final palindromic string which is // lexicographically first return (front_str + odd_str + rear_str); } // Driver program public static void main(String[] args) { String str = "malayalam"; System.out.println(findPalindromicString(str)); } } // This code is contributed by Rajput-JiPython3
# Python3 program to find first palindromic permutation # of given string MAX_CHAR = 26; # Function to count frequency of each char in the # string. freq[0] for 'a',...., freq[25] for 'z' def countFreq(str1, freq, len1): for i in range(len1): freq[ord(str1[i]) - ord('a')] += 1; # Cases to check whether a palindr0mic # string can be formed or not def canMakePalindrome(freq, len1): # count_odd to count no of # chars with odd frequency count_odd = 0; for i in range(MAX_CHAR): if (freq[i] % 2 != 0): count_odd += 1; # For even length string # no odd freq character if (len1 % 2 == 0): if (count_odd > 0): return False; else: return True; # For odd length string # one odd freq character if (count_odd != 1): return False; return True; # Function to find odd freq char and # reducing its freq by 1returns "" if odd freq # char is not present def findOddAndRemoveItsFreq(freq): odd_str = ""; for i in range(MAX_CHAR): if (freq[i]%2 != 0): freq[i]-=1; odd_str += chr(i+ord('a')); return odd_str; return odd_str; # To find lexicographically first palindromic # string. def findPalindromicString(str1): len1 = len(str1); freq=[0]*MAX_CHAR; countFreq(str1, freq, len1); if (canMakePalindrome(freq, len1) == False): return "No Palindromic String"; # Assigning odd freq character if present # else empty string. odd_str = findOddAndRemoveItsFreq(freq); front_str = ""; rear_str = " "; # Traverse characters in increasing order for i in range(MAX_CHAR): temp = ""; if (freq[i] != 0): ch = chr(i + ord('a')); # Divide all occurrences into two # halves. Note that odd character # is removed by findOddAndRemoveItsFreq() for j in range(1,int(freq[i]/2)+1): temp += ch; # creating front string front_str += temp; # creating rear string rear_str = temp+rear_str; # Final palindromic string which is # lexicographically first return (front_str + odd_str+rear_str); # Driver code str1 = "malayalam"; print(findPalindromicString(str1)); # This code is contributed by mitsC#
// C# program to find first palindromic permutation // of given string using System; class GFG { static int MAX_CHAR = 26; // Function to count frequency of each char in the // string. freq[0] for 'a',...., freq[25] for 'z' static void countFreq(string str, int[] freq, int len) { for (int i = 0; i < len; i++) { freq[str[i] - 'a']++; } } // Cases to check whether a palindr0mic // string can be formed or not static bool canMakePalindrome(int[] freq, int len) { // count_odd to count no of // chars with odd frequency int count_odd = 0; for (int i = 0; i < MAX_CHAR; i++) { if (freq[i] % 2 != 0) { count_odd++; } } // For even length string // no odd freq character if (len % 2 == 0) { if (count_odd > 0) { return false; } else { return true; } } // For odd length string // one odd freq character if (count_odd != 1) { return false; } return true; } // Function to find odd freq char and // reducing its freq by 1returns "" if odd freq // char is not present static string findOddAndRemoveItsFreq(int[] freq) { string odd_str = ""; for (int i = 0; i < MAX_CHAR; i++) { if (freq[i] % 2 != 0) { freq[i]--; odd_str = odd_str + (char) (i + 'a'); return odd_str; } } return odd_str; } // To find lexicographically first // palindromic string. static string findPalindromicString(string str) { int len = str.Length; int[] freq = new int[MAX_CHAR]; countFreq(str, freq, len); if (!canMakePalindrome(freq, len)) { return "No Palindromic String"; } // Assigning odd freq character if present // else empty string. string odd_str = findOddAndRemoveItsFreq(freq); string front_str = "", rear_str = " "; // Traverse characters in increasing order for (int i = 0; i < MAX_CHAR; i++) { String temp = ""; if (freq[i] != 0) { char ch = (char) (i + 'a'); // Divide all occurrences into two // halves. Note that odd character // is removed by findOddAndRemoveItsFreq() for (int j = 1; j <= freq[i] / 2; j++) { temp = temp + ch; } // creating front string front_str = front_str + temp; // creating rear string rear_str = temp + rear_str; } } // Final palindromic string which is // lexicographically first return (front_str + odd_str + rear_str); } // Driver code public static void Main() { string str = "malayalam"; Console.Write(findPalindromicString(str)); } } // This code is contributed by Ita_c.PHP
0) return false; else return true; } // For odd length string // one odd freq character if ($count_odd != 1) return false; return true; } // Function to find odd freq char and // reducing its freq by 1returns "" if odd freq // char is not present function findOddAndRemoveItsFreq($freq) { global $MAX_CHAR; $odd_str = ""; for ($i = 0; $i < $MAX_CHAR; $i++) { if ($freq[$i] % 2 != 0) { $freq[$i]--; $odd_str .= chr($i+ord('a')); return $odd_str; } } return $odd_str; } // To find lexicographically first palindromic // string. function findPalindromicString($str) { global $MAX_CHAR; $len = strlen($str); $freq=array_fill(0, $MAX_CHAR, 0); countFreq($str, $freq, $len); if (!canMakePalindrome($freq, $len)) return "No Palindromic String"; // Assigning odd freq character if present // else empty string. $odd_str = findOddAndRemoveItsFreq($freq); $front_str = ""; $rear_str = " "; // Traverse characters in increasing order for ($i = 0; $i < $MAX_CHAR; $i++) { $temp = ""; if ($freq[$i] != 0) { $ch = chr($i + ord('a')); // Divide all occurrences into two // halves. Note that odd character // is removed by findOddAndRemoveItsFreq() for ($j = 1; $j <= (int)($freq[$i]/2); $j++) $temp .= $ch; // creating front string $front_str .= $temp; // creating rear string $rear_str = $temp.$rear_str; } } // Final palindromic string which is // lexicographically first return ($front_str.$odd_str.$rear_str); } // Driver code $str = "malayalam"; echo findPalindromicString($str); // This code is contributed by mits ?>输出：
aalmymlaa时间复杂度： O(n) 其中 n 是输入字符串的长度。假设字符串字母的大小是恒定的。