生成字符串回文所需的最小附加数
给定一个字符串s，我们需要告诉最少要附加的字符（在末尾插入）以形成一个字符串回文。
例子：
Input : s = "abede"
Output : 2
We can make string palindrome as "abedeba"
by adding ba at the end of the string.
Input : s = "aabb"
Output : 2
We can make string palindrome as"aabbaa"
by adding aa at the end of the string.可以通过从字符串的开头一个一个地删除字符并检查该字符串是否为回文来实现该解决方案。
例如，考虑上面的字符串s = “abede” 。
我们检查字符串是否为回文。
结果是假的，然后我们从字符串的开头删除字符，现在字符串变成了“bede” 。
我们检查字符串是否为回文。结果再次为假，然后我们从字符串的开头删除字符，现在字符串变为“ede” 。
我们检查字符串是否为回文。结果为真，因此输出变为 2，即从字符串中删除的字符数。
C++
// C program to find minimum number of appends
// needed to make a string Palindrome
#include
#include
#include
// Checking if the string is palindrome or not
bool isPalindrome(char *str)
{
int len = strlen(str);
// single character is always palindrome
if (len == 1)
return true;
// pointing to first character
char *ptr1 = str;
// pointing to last character
char *ptr2 = str+len-1;
while (ptr2 > ptr1)
{
if (*ptr1 != *ptr2)
return false;
ptr1++;
ptr2--;
}
return true;
}
// Recursive function to count number of appends
int noOfAppends(char s[])
{
if (isPalindrome(s))
return 0;
// Removing first character of string by
// incrementing base address pointer.
s++;
return 1 + noOfAppends(s);
}
// Driver program to test above functions
int main()
{
char s[] = "abede";
printf("%d\n", noOfAppends(s));
return 0;
} Java
// Java program to find minimum number of appends
// needed to make a string Palindrome
class GFG
{
// Checking if the string is palindrome or not
static boolean isPalindrome(char []str)
{
int len = str.length;
// single character is always palindrome
if (len == 1)
return true;
// pointing to first character
int ptr1 = 0;
// pointing to last character
int ptr2 = len-1;
while (ptr2 >= ptr1)
{
if (str[ptr1] != str[ptr2])
return false;
ptr1++;
ptr2--;
}
return true;
}
// Recursive function to count number of appends
static int noOfAppends(String s)
{
if (isPalindrome(s.toCharArray()))
return 0;
// Removing first character of string by
// incrementing base address pointer.
s=s.substring(1);
return 1 + noOfAppends(s);
}
// Driver code
public static void main(String arr[])
{
String s = "abede";
System.out.printf("%d\n", noOfAppends(s));
}
}
// This code contributed by Rajput-JiPython3
# Python3 program to find minimum number of appends
# needed to make a String Palindrome
# Checking if the String is palindrome or not
def isPalindrome(Str):
Len = len(Str)
# single character is always palindrome
if (Len == 1):
return True
# pointing to first character
ptr1 = 0
# pointing to last character
ptr2 = Len - 1
while (ptr2 > ptr1):
if (Str[ptr1] != Str[ptr2]):
return False
ptr1 += 1
ptr2 -= 1
return True
# Recursive function to count number of appends
def noOfAppends(s):
if (isPalindrome(s)):
return 0
# Removing first character of String by
# incrementing base address pointer.
del s[0]
return 1 + noOfAppends(s)
# Driver Code
se = "abede"
s = [i for i in se]
print(noOfAppends(s))
# This code is contributed by Mohit KumarC#
// C# program to find minimum number of appends
// needed to make a string Palindrome
using System;
class GFG
{
// Checking if the string is palindrome or not
static Boolean isPalindrome(char []str)
{
int len = str.Length;
// single character is always palindrome
if (len == 1)
return true;
// pointing to first character
char ptr1 = str[0];
// pointing to last character
char ptr2 = str[len-1];
while (ptr2 > ptr1)
{
if (ptr1 != ptr2)
return false;
ptr1++;
ptr2--;
}
return true;
}
// Recursive function to count number of appends
static int noOfAppends(String s)
{
if (isPalindrome(s.ToCharArray()))
return 0;
// Removing first character of string by
// incrementing base address pointer.
s=s.Substring(1);
return 1 + noOfAppends(s);
}
// Driver code
public static void Main(String []arr)
{
String s = "abede";
Console.Write("{0}\n", noOfAppends(s));
}
}
// This code has been contributed by 29AjayKumarJavascript
C++
// CPP program for above approach
#include
#include
#include
using namespace std;
// This class builds the dfa and
// precomputes the state.
// See KMP algorithm for explanation
class kmp_numeric {
private:
int n;
int** dfa;
public:
kmp_numeric(string& s)
{
n = s.length();
int c = 256;
// Create dfa
dfa = new int*[n];
// Iterate from 0 to n
for (int i = 0; i < n; i++)
dfa[i] = new int;
int x = 0;
// Iterate from 0 to n
for (int i = 0; i < c; i++)
dfa[0][i] = 0;
// Initialise dfa[0][s[0]] = 1
dfa[0][s[0]] = 1;
// Iterate i from 1 to n-1
for (int i = 1; i < n; i++) {
// Iterate j from 0 to c - 1
for (int j = 0; j < c; j++) {
dfa[i][j] = dfa[x][j];
}
dfa[i][s[i]] = i + 1;
x = dfa[x][s[i]];
}
}
// This function finds the overlap
// between two strings,by
// changing the state.
int longest_overlap(string& query)
{
// q1 is length of query
int ql = query.length();
int state = 0;
// Iterate from 0 to q1 - 1
for (int i = 0; i < ql; i++) {
state = dfa[state][query[i]];
}
return state;
}
};
int min_appends(string& s)
{
// Reverse the string.
reverse(s.begin(), s.end());
// Build the DFA for the
// reversed String
kmp_numeric kmp = s;
// Get the original string back
reverse(s.begin(), s.end());
// Largest overlap in this case is the
// largest string from the end which
// is a palindrome.
int ans = s.length() - kmp.longest_overlap(s);
return ans;
}
// Driver Code
int main()
{
string s = "deep";
// Answer : 3
string t = "sososososos";
// Answer : 0
cout << min_appends(s) << endl;
cout << min_appends(t) << endl;
} 输出
2描述了上述方法和 O(n**2) 方法。
有效的方法：
我们还有一个借助 Knuth Morris Pratt 算法的算法，它是 O(n) 时间复杂度。
该方法背后的基本思想是，我们从末尾计算最大的子字符串，可以计算出字符串的长度减去这个值是追加的最小数量。逻辑很直观，我们不需要附加回文，只需要那些不形成回文的。为了从末尾找到这个最大的回文串，我们反转字符串，计算 DFA 并再次反转字符串（从而获得原始字符串）并找到最终状态，它表示字符串与被尊崇的字符串的匹配次数和因此，我们在 O(n) 时间内从末尾得到最大的回文子串。
下面是上述方法的实现：
C++
// CPP program for above approach
#include
#include
#include
using namespace std;
// This class builds the dfa and
// precomputes the state.
// See KMP algorithm for explanation
class kmp_numeric {
private:
int n;
int** dfa;
public:
kmp_numeric(string& s)
{
n = s.length();
int c = 256;
// Create dfa
dfa = new int*[n];
// Iterate from 0 to n
for (int i = 0; i < n; i++)
dfa[i] = new int;
int x = 0;
// Iterate from 0 to n
for (int i = 0; i < c; i++)
dfa[0][i] = 0;
// Initialise dfa[0][s[0]] = 1
dfa[0][s[0]] = 1;
// Iterate i from 1 to n-1
for (int i = 1; i < n; i++) {
// Iterate j from 0 to c - 1
for (int j = 0; j < c; j++) {
dfa[i][j] = dfa[x][j];
}
dfa[i][s[i]] = i + 1;
x = dfa[x][s[i]];
}
}
// This function finds the overlap
// between two strings,by
// changing the state.
int longest_overlap(string& query)
{
// q1 is length of query
int ql = query.length();
int state = 0;
// Iterate from 0 to q1 - 1
for (int i = 0; i < ql; i++) {
state = dfa[state][query[i]];
}
return state;
}
};
int min_appends(string& s)
{
// Reverse the string.
reverse(s.begin(), s.end());
// Build the DFA for the
// reversed String
kmp_numeric kmp = s;
// Get the original string back
reverse(s.begin(), s.end());
// Largest overlap in this case is the
// largest string from the end which
// is a palindrome.
int ans = s.length() - kmp.longest_overlap(s);
return ans;
}
// Driver Code
int main()
{
string s = "deep";
// Answer : 3
string t = "sososososos";
// Answer : 0
cout << min_appends(s) << endl;
cout << min_appends(t) << endl;
}
输出
3
0建议： Pratik Priyadarsan