For example the input is 'abba' then the possible palindromes= a, b, b, a, bb, abba
So the result is 6.
Updated at 11/11/2013:
After the interview I got know that the O(n^3) solution is not enough to go to the next round. It would have been better to know before starting implementing the solution unnecessarily ...
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Similar to leetcode Longest Palindromic Substring Part II in my blog, the code is like:
#include <iostream>
#include <map>
#include <algorithm>
#include <limits.h>
#include <assert.h>
#include <string.h>
#include <vector>
using namespace std;
string preprocess(string s) {
string res = "^#";
for (int i = 0; i < s.length(); ++i) {
res += s[i];
res += '#';
}
res += '$';
return res;
}
int getPalindromeNum(string s) {
string str = preprocess(s);
int i, j, len = str.length(), C = 0, R = 0, res = 0, ii;
vector<int> T(len + 1, 0), P(len + 1, 0);
for (i = 1; i < len; ++i) {
ii = 2*C - i;
P[i] = (R - i) > 0 ?
min(P[ii], R-i) : 0; //bug1: P[i] = (R - i) > 0 ? P[i] : 0 while (str[i + P[i] + 1] == str[i - P[i] - 1]) ++P[i]; res += (P[i] + 1) / 2; //bug2: res += P[i]; if (i + P[i] > R) { C = i; R = i + P[i]; } } return res; } int main() { //string s = "abcba"; string s = "aaaaa"; int res = getPalindromeNum(s); return 0; }