形成字符串回文的逆时针移位次数

给定字符串小写英文字母，找出使字符串回文所需的字符逆时针移位次数。假定移动字符串总是会导致回文。
例子：

Input: str = “baabbccb” 
Output: 2 
Shifting the string counterclockwise 2 times, 
will make the string palindrome. 
1st shift : aabbccbb 
2nd shift :abbccbba

Input: bbaabbcc 
Output: 3 
Shifting the string counterclockwise 
3 times will make the string palindrome. 
1st shift : baabbccb 
2nd shift : aabbccbb 
3rd shift : abbccbba

朴素的方法：朴素的方法是将给定字符串的一个字符一个一个地逆时针循环，并检查该字符串是否为回文。

更好的方法：更好的方法是将字符串附加到自身并从给定字符串的第一个字符迭代到最后一个字符。附加字符串从 i 到 i+n（其中 i 在 [0, n-1] 范围内）的子字符串将是每次逆时针移动后获得的字符串。检查子串是否为回文。移位操作的次数将为 i。

下面是上述方法的实现：

C++

// C++ program to find counter clockwise
// shifts to make string palindrome.
#include 
using namespace std;
 
// Function to check if given string is
// palindrome or not.
bool isPalindrome(string str, int l, int r)
{
    while (l < r) {
        if (str[l] != str[r])
            return false;
 
        l++;
        r--;
    }
 
    return true;
}
 
// Function to find counter clockwise shifts
// to make string palindrome.
int CyclicShifts(string str)
{
 
    int n = str.length();
 
    // Pointer to starting of current
    // shifted string.
    int left = 0;
 
    // Pointer to ending of current
    // shifted string.
    int right = n - 1;
 
    // Concatenate string with itself
    str = str + str;
 
    // To store counterclockwise shifts
    int cnt = 0;
 
    // Move left and right pointers one
    // step at a time.
    while (right < 2 * n - 1) {
 
        // Check if current shifted string
        // is palindrome or not
        if (isPalindrome(str, left, right))
            break;
 
        // If string is not palindrome
        // then increase count of number
        // of shifts by 1.
        cnt++;
 
        left++;
        right++;
    }
 
    return cnt;
}
 
// Driver code.
int main()
{
    string str = "bccbbaab";
 
    cout << CyclicShifts(str);
    return 0;
}

Java

// Java program to find counter clockwise
// shifts to make string palindrome.
class GFG {
 
    // Function to check if given string is
    // palindrome or not.
    static boolean isPalindrome(String str, int l, int r)
    {
        while (l < r) {
            if (str.charAt(l) != str.charAt(r))
                return false;
 
            l++;
            r--;
        }
        return true;
    }
 
    // Function to find counter clockwise shifts
    // to make string palindrome.
    static int CyclicShifts(String str)
    {
 
        int n = str.length();
 
        // Pointer to starting of current
        // shifted string.
        int left = 0;
 
        // Pointer to ending of current
        // shifted string.
        int right = n - 1;
 
        // Concatenate string with itself
        str = str + str;
 
        // To store counterclockwise shifts
        int cnt = 0;
 
        // Move left and right pointers one
        // step at a time.
        while (right < 2 * n - 1) {
 
            // Check if current shifted string
            // is palindrome or not
            if (isPalindrome(str, left, right))
                break;
 
            // If string is not palindrome
            // then increase count of number
            // of shifts by 1.
            cnt++;
 
            left++;
            right++;
        }
        return cnt;
    }
 
    // Driver code.
    public static void main(String[] args)
    {
        String str = "bccbbaab";
 
        System.out.println(CyclicShifts(str));
    }
}
 
// This code is contributed by Rajput-Ji

Python3

# Python3 program to find counter clockwise
# shifts to make string palindrome.
 
# Function to check if given string
# is palindrome or not.
def isPalindrome(str, l, r):
 
    while (l < r) :
        if (str[l] != str[r]):
            return False
 
        l += 1
        r -= 1
 
    return True
 
# Function to find counter clockwise
# shifts to make string palindrome.
def CyclicShifts(str):
 
    n = len(str)
 
    # Pointer to starting of current
    # shifted string.
    left = 0
 
    # Pointer to ending of current
    # shifted string.
    right = n - 1
 
    # Concatenate string with itself
    str = str + str
 
    # To store counterclockwise shifts
    cnt = 0
 
    # Move left and right pointers
    # one step at a time.
    while (right < 2 * n - 1) :
 
        # Check if current shifted string
        # is palindrome or not
        if (isPalindrome(str, left, right)):
            break
 
        # If string is not palindrome
        # then increase count of number
        # of shifts by 1.
        cnt += 1
 
        left += 1
        right += 1
 
    return cnt
 
# Driver code.
if __name__ == "__main__":
     
    str = "bccbbaab";
 
    print(CyclicShifts(str))
 
# This code is contributed by ita_c

C#

// C# program to find counter clockwise
// shifts to make string palindrome.
using System;
 
class GFG
{
 
    // Function to check if given string is
    // palindrome or not.
    static bool isPalindrome(String str, int l, int r)
    {
        while (l < r)
        {
            if (str[l] != str[r])
                return false;
 
            l++;
            r--;
        }
        return true;
    }
 
    // Function to find counter clockwise shifts
    // to make string palindrome.
    static int CyclicShifts(String str)
    {
 
        int n = str.Length;
 
        // Pointer to starting of current
        // shifted string.
        int left = 0;
 
        // Pointer to ending of current
        // shifted string.
        int right = n - 1;
 
        // Concatenate string with itself
        str = str + str;
 
        // To store counterclockwise shifts
        int cnt = 0;
 
        // Move left and right pointers one
        // step at a time.
        while (right < 2 * n - 1)
        {
 
            // Check if current shifted string
            // is palindrome or not
            if (isPalindrome(str, left, right))
                break;
 
            // If string is not palindrome
            // then increase count of number
            // of shifts by 1.
            cnt++;
 
            left++;
            right++;
        }
        return cnt;
    }
 
    // Driver code.
    public static void Main(String[] args)
    {
        String str = "bccbbaab";
 
        Console.WriteLine(CyclicShifts(str));
    }
}
 
// This code is contributed by 29AjayKumar

Javascript

 
 
// Javascript program to find counter clockwise
// shifts to make string palindrome.
 
// Function to check if given string is
// palindrome or not.
function isPalindrome(str, l, r)
{
    while (l < r) {
        if (str[l] != str[r])
            return false;
 
        l++;
        r--;
    }
 
    return true;
}
 
// Function to find counter clockwise shifts
// to make string palindrome.
function CyclicShifts(str)
{
 
    var n = str.length;
 
    // Pointer to starting of current
    // shifted string.
    var left = 0;
 
    // Pointer to ending of current
    // shifted string.
    var right = n - 1;
 
    // Concatenate string with itself
    str = str + str;
 
    // To store counterclockwise shifts
    var cnt = 0;
 
    // Move left and right pointers one
    // step at a time.
    while (right < 2 * n - 1) {
 
        // Check if current shifted string
        // is palindrome or not
        if (isPalindrome(str, left, right))
            break;
 
        // If string is not palindrome
        // then increase count of number
        // of shifts by 1.
        cnt++;
 
        left++;
        right++;
    }
 
    return cnt;
}
 
// Driver code.
var str = "bccbbaab";
document.write(CyclicShifts(str));
 
 

输出：

2

时间复杂度： O(N ² )
辅助空间： O(N)

高效方法：一种有效的方法是使用累积哈希。该字符串根据上述方法循环移位，并将该字符串的哈希值与反向字符串的哈希值进行比较。如果两个值相同，则当前移位的字符串是回文，否则字符串再次移位。在任何一步，班次计数都是 i 。使用哈希函数计算两个字符串的值：

H(s) = ∑ (31ⁱ * (S_i – ‘a’)) % mod, 0 ≤ i ≤ (length of string – 1) 
where, H(x) = Hash function 
s = given string 
mod = 10⁹ + 7 

迭代所有子串并使用上述散列函数和累积散列技术检查它是否是回文。

下面是上述方法的实现：

C++

// CPP program to find counter clockwise
// shifts to make string palindrome.
#include 
 
#define mod 1000000007
using namespace std;
 
// Function that returns true
// if str is palindrome
bool isPalindrome(string str, int n)
{
    int i = 0, j = n - 1;
    while (i < j) {
        if (str[i] != str[j])
            return false;
        i++;
        j--;
    }
    return true;
}
 
// Function to find counter clockwise shifts
// to make string palindrome.
int CyclicShifts(string str)
{
 
    int n = str.length(), i;
 
    // If the string is already a palindrome
    if (isPalindrome(str, n))
        return 0;
 
    // To store power of 31.
    // po[i] = 31^i;
    long long int po[2 * n + 2];
 
    // To store hash value of string.
    long long int preval[2 * n + 2];
 
    // To store hash value of reversed
    // string.
    long long int suffval[2 * n + 2];
 
    // To find hash value of string str[i..j]
    long long int val1;
 
    // To store hash value of reversed string
    // str[j..i]
    long long int val2;
 
    // To store number of counter clockwise
    // shifts.
    int cnt = 0;
 
    // Concatenate string with itself to shift
    // it cyclically.
    str = str + str;
 
    // Calculate powers of 31 upto 2*n which
    // will be used in hash function.
    po[0] = 1;
    for (i = 1; i <= 2 * n; i++) {
        po[i] = (po[i - 1] * 31) % mod;
    }
 
    // Hash value of string str[0..i] is stored in
    // preval[i].
    for (i = 1; i <= 2 * n; i++) {
        preval[i] = ((preval[i - 1] * 31) % mod + (str[i - 1] - 'a')) % mod;
    }
 
    // Hash value of string str[i..n-1] is stored
    // in suffval[i].
    for (i = 2 * n; i > 0; i--) {
        suffval[i] = ((suffval[i + 1] * 31) % mod + (str[i - 1] - 'a')) % mod;
    }
 
    // Characters in string str[0..i] is present
    // at position [(n-1-i)..(n-1)] in reversed
    // string. If hash value of both are same
    // then string is palindrome else not.
    for (i = 1; i <= n; i++) {
 
        // Hash value of shifted string starting at
        // index i and ending at index i+n-1.
        val1 = (preval[i + n - 1] - ((po[n] * preval[i - 1]) % mod)) % mod;
        if (val1 < 0)
            val1 += mod;
 
        // Hash value of corresponding string when
        // reversed starting at index i+n-1 and
        // ending at index i.
        val2 = (suffval[i] - ((po[n] * suffval[i + n])
                              % mod))
               % mod;
        if (val2 < 0)
            val2 += mod;
 
        // If both hash value are same then current
        // string str[i..(i+n-1)] is palindrome.
        // Else increase the shift count.
        if (val1 != val2)
            cnt++;
        else
            break;
    }
 
    return cnt;
}
 
// Driver code.
int main()
{
    string str = "bccbbaab";
 
    cout << CyclicShifts(str);
    return 0;
}

Java

// Java program to find counter clockwise
// shifts to make string palindrome.
class GFG{
     
static int mod = 1000000007;
 
// Function that returns true
// if str is palindrome
public static boolean isPalindrome(String str, int n)
{
    int i = 0, j = n - 1;
     
    while (i < j)
    {
        if (str.charAt(i) != str.charAt(j))
            return false;
             
        i++;
        j--;
    }
    return true;
}
    
// Function to find counter clockwise shifts
// to make string palindrome.
public static int CyclicShifts(String str)
{
    int n = str.length(), i;
    
    // If the string is already a palindrome
    if (isPalindrome(str, n))
        return 0;
    
    // To store power of 31.
    // po[i] = 31^i;
    long[] po = new long[2 * n + 2];
    
    // To store hash value of string.
    long[] preval = new long[2 * n + 2];
    
    // To store hash value of reversed
    // string.
    long[] suffval = new long[2 * n + 2];
    
    // To find hash value of string str[i..j]
    long val1;
    
    // To store hash value of reversed string
    // str[j..i]
    long val2;
    
    // To store number of counter clockwise
    // shifts.
    int cnt = 0;
    
    // Concatenate string with itself to shift
    // it cyclically.
    str = str + str;
    
    // Calculate powers of 31 upto 2*n which
    // will be used in hash function.
    po[0] = 1;
     
    for(i = 1; i <= 2 * n; i++)
    {
        po[i] = (po[i - 1] * 31) % mod;
    }
    
    // Hash value of string str[0..i] is stored in
    // preval[i].
    for(i = 1; i <= 2 * n; i++)
    {
        preval[i] = ((preval[i - 1] * 31) % mod +
                 (str.charAt(i - 1) - 'a')) % mod;
    }
    
    // Hash value of string str[i..n-1] is stored
    // in suffval[i].
    for(i = 2 * n; i > 0; i--)
    {
        suffval[i] = ((suffval[i + 1] * 31) % mod +
                   (str.charAt(i - 1) - 'a')) % mod;
    }
    
    // Characters in string str[0..i] is present
    // at position [(n-1-i)..(n-1)] in reversed
    // string. If hash value of both are same
    // then string is palindrome else not.
    for(i = 1; i <= n; i++)
    {
         
        // Hash value of shifted string starting at
        // index i and ending at index i+n-1.
        val1 = (preval[i + n - 1] -
                  ((po[n] *
                preval[i - 1]) % mod)) % mod;
                 
        if (val1 < 0)
            val1 += mod;
    
        // Hash value of corresponding string when
        // reversed starting at index i+n-1 and
        // ending at index i.
        val2 = (suffval[i] -
                   ((po[n] *
                suffval[i + n]) % mod)) % mod;
                 
        if (val2 < 0)
            val2 += mod;
    
        // If both hash value are same then current
        // string str[i..(i+n-1)] is palindrome.
        // Else increase the shift count.
        if (val1 != val2)
            cnt++;
        else
            break;
    }
    return cnt;
}
 
// Driver code
public static void main(String[] args)
{
    String str = "bccbbaab";
 
    System.out.println(CyclicShifts(str));
}
}
 
// This code is contributed by divyeshrabadiya07

蟒蛇3

# Python3 program to find counter clockwise
# shifts to make string palindrome.
mod = 1000000007
 
# Function to find counter clockwise shifts
# to make string palindrome.
def CyclicShifts(str1):
 
    n = len(str1)
    i = 0
 
    # To store power of 31.
    # po[i] = 31 ^ i;
    po = [0 for i in range(2 * n + 2)]
 
    # To store hash value of string.
    preval = [0 for i in range(2 * n + 2)]
 
    # To store hash value of reversed
    # string.
    suffval = [0 for i in range(2 * n + 2)]
 
    # To find hash value of string str[i..j]
    val1 = 0
 
    # To store hash value of reversed string
    # str[j..i]
    val2 = 0
 
    # To store number of counter clockwise
    # shifts.
    cnt = 0
 
    # Concatenate string with itself to shift
    # it cyclically.
    str1 = str1 + str1
 
    # Calculate powers of 31 upto 2 * n which
    # will be used in hash function.
    po[0] = 1
    for i in range(1, 2 * n + 1):
        po[i] = (po[i - 1] * 31) % mod
 
    # Hash value of string str[0..i]
    # is stored in preval[i].
    for i in range(1, 2 * n + 1):
        preval[i] = ((preval[i - 1] * 31) % mod +
                        (ord(str1[i - 1]) -
                         ord('a'))) % mod
     
    # Hash value of string str[i..n-1] is stored
    # in suffval[i].
    for i in range(2 * n, -1, -1):
        suffval[i] = ((suffval[i + 1] * 31) % mod +
                      (ord(str1[i - 1]) -
                       ord('a'))) % mod
 
    # Characters in string str[0..i] is present
    # at position [(n-1-i)..(n-1)] in reversed
    # string. If hash value of both are same
    # then string is palindrome else not.
    for i in range(1, n + 1):
 
        # Hash value of shifted string starting at
        # index i and ending at index i + n-1.
        val1 = (preval[i + n - 1] - ((po[n] *
                preval[i - 1]) % mod)) % mod
        if (val1 < 0):
            val1 += mod
 
        # Hash value of corresponding string when
        # reversed starting at index i + n-1 and
        # ending at index i.
        val2 = (suffval[i] - ((po[n] *
                suffval[i + n])% mod)) % mod;
        if (val2 < 0):
            val2 += mod
 
        # If both hash value are same then current
        # string str[i..(i + n-1)] is palindrome.
        # Else increase the shift count.
        if (val1 != val2):
            cnt += 1
        else:
            break
     
    return cnt
 
# Driver code
str1 = "bccbbaab"
 
print(CyclicShifts(str1))
 
# This code is contributed by mohit kumar

C＃

// C# program to find counter clockwise
// shifts to make string palindrome.
using System;
using System.Collections.Generic;
 
class GFG
{
     
static int mod= 1000000007;
 
// Function that returns true
// if str is palindrome
static bool isPalindrome(string str, int n)
{
    int i = 0, j = n - 1;
    while (i < j) {
        if (str[i] != str[j])
            return false;
        i++;
        j--;
    }
    return true;
}
  
// Function to find counter clockwise shifts
// to make string palindrome.
static int CyclicShifts(string str)
{
  
    int n = str.Length, i;
  
    // If the string is already a palindrome
    if (isPalindrome(str, n))
        return 0;
  
    // To store power of 31.
    // po[i] = 31^i;
    long []po=new long[2 * n + 2];
  
    // To store hash value of string.
    long []preval=new long[2 * n + 2];
  
    // To store hash value of reversed
    // string.
    long []suffval=new long[2 * n + 2];
  
    // To find hash value of string str[i..j]
    long val1;
  
    // To store hash value of reversed string
    // str[j..i]
    long val2;
  
    // To store number of counter clockwise
    // shifts.
    int cnt = 0;
  
    // Concatenate string with itself to shift
    // it cyclically.
    str = str + str;
  
    // Calculate powers of 31 upto 2*n which
    // will be used in hash function.
    po[0] = 1;
    for (i = 1; i <= 2 * n; i++) {
        po[i] = (po[i - 1] * 31) % mod;
    }
  
    // Hash value of string str[0..i] is stored in
    // preval[i].
    for (i = 1; i <= 2 * n; i++) {
        preval[i] = ((preval[i - 1] * 31) % mod + (str[i - 1] - 'a')) % mod;
    }
  
    // Hash value of string str[i..n-1] is stored
    // in suffval[i].
    for (i = 2 * n; i > 0; i--) {
        suffval[i] = ((suffval[i + 1] * 31) % mod + (str[i - 1] - 'a')) % mod;
    }
  
    // Characters in string str[0..i] is present
    // at position [(n-1-i)..(n-1)] in reversed
    // string. If hash value of both are same
    // then string is palindrome else not.
    for (i = 1; i <= n; i++) {
  
        // Hash value of shifted string starting at
        // index i and ending at index i+n-1.
        val1 = (preval[i + n - 1] - ((po[n] * preval[i - 1]) % mod)) % mod;
        if (val1 < 0)
            val1 += mod;
  
        // Hash value of corresponding string when
        // reversed starting at index i+n-1 and
        // ending at index i.
        val2 = (suffval[i] - ((po[n] * suffval[i + n])
                              % mod))
               % mod;
        if (val2 < 0)
            val2 += mod;
  
        // If both hash value are same then current
        // string str[i..(i+n-1)] is palindrome.
        // Else increase the shift count.
        if (val1 != val2)
            cnt++;
        else
            break;
    }
  
    return cnt;
}
       
    // Driver Code
    public static void Main(string []args)
    {
         string str = "bccbbaab";
  
        Console.Write(CyclicShifts(str));
    }
}

输出：

2

时间复杂度： O(N)
辅助空间： O(N)

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