Created
September 16, 2012 09:24
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find the largest substring palindrome in a given string in linear time
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#include<iostream.h> | |
#include<string.h> | |
// Transform S into T. | |
// For example, S = "abba", T = "^#a#b#b#a#$". | |
// ^ and $ signs are sentinels appended to each end to avoid bounds checking | |
string preProcess(string s) { | |
int n = s.length(); | |
if (n == 0) return "^$"; | |
string ret = "^"; | |
for (int i = 0; i < n; i++) | |
ret += "#" + s.substr(i, 1); | |
ret += "#$"; | |
return ret; | |
} | |
string longestPalindrome(string s) { | |
string T = preProcess(s); | |
int n = T.length(); | |
int *P = new int[n]; | |
int C = 0, R = 0; | |
for (int i = 1; i < n-1; i++) { | |
int i_mirror = 2*C-i; // equals to i' = C - (i-C) | |
P[i] = (R > i) ? min(R-i, P[i_mirror]) : 0; | |
// Attempt to expand palindrome centered at i | |
while (T[i + 1 + P[i]] == T[i - 1 - P[i]]) | |
P[i]++; | |
// If palindrome centered at i expand past R, | |
// adjust center based on expanded palindrome. | |
if (i + P[i] > R) { | |
C = i; | |
R = i + P[i]; | |
} | |
} | |
// Find the maximum element in P. | |
int maxLen = 0; | |
int centerIndex = 0; | |
for (int i = 1; i < n-1; i++) { | |
if (P[i] > maxLen) { | |
maxLen = P[i]; | |
centerIndex = i; | |
} | |
} | |
delete[] P; | |
return s.substr((centerIndex - 1 - maxLen)/2, maxLen); | |
} | |
int main() | |
{ | |
string a="racecarmalayalam"; | |
string output; | |
output=longestPalindrome(a); | |
cout<<output; | |
} |
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