/**
* Longest Palindromic Substring
* 最長迴文子串
*/
#include <string.h>
#include <iostream>
#include <string>
#include <vector>
using namespace std;
string findPalindrome(string s, int left, int right)
{
int n = s.size();
int l = left;
int r = right;
while (left>=0 && right<=n-1 && s[left] == s[right]) {
left--;
right++;
}
return s.substr(left+1, right-left-1);
}
// This is the common solution.
// Actuatlly it's faster than DP solution under Leetcode's test
// the below optimized DP solution need 700ms+, this needs around 250ms.
string longestPalindrome_recursive_way(string s) {
int n = s.size();
if (n<=1) return s;
string longest;
string str;
for (int i=0; i<n-1; i++) {
str = findPalindrome(s, i, i);
if (str.size() > longest.size()){
longest = str;
}
str = findPalindrome(s, i, i+1);
if (str.size() > longest.size()){
longest = str;
}
}
return longest;
}
//================================================================================
void findPalindrome(string s, int left, int right, int& start, int& len)
{
int n = s.size();
int l = left;
int r = right;
while (left>=0 && right<=n-1 && s[left] == s[right]) {
left--;
right++;
}
if (right-left-1 > len){
len = right-left-1;
start = left+1;
}
}
//The following solution is better than previous solution.
//Because it remove the sub-string return in findPalindrome().
string longestPalindrome_recursive_way2(string s) {
int n = s.size();
if (n<=1) return s;
int start=0, len=0;
string longest;
string str;
for (int i=0; i<n-1; i++) {
findPalindrome(s, i, i, start, len);
findPalindrome(s, i, i+1, start, len);
}
return s.substr(start, len);
}
//================================================================================
// Time/Memory Limit Exceeded
string longestPalindrome_dp_way(string s) {
string longest;
int n = s.size();
if (n<=1) return s;
//Construct a matrix, and consdier matrix[i][j] as s[i] -> s[j] is Palindrome or not.
//using char or int could cause the `Memory Limit Error`
//vector< vector<char> > matrix (n, vector<char>(n));
//using bool type could cause the `Time Limit Error`
vector< vector<bool> > matrix (n, vector<bool>(n));
// Dynamic Programming
// 1) if i == j, then matrix[i][j] = true;
// 2) if i != j, then matrix[i][j] = (s[i]==s[j] && matrix[i+1][j-1])
for (int i=n-1; i>=0; i--){
for (int j=i; j<n; j++){
// The following if statement can be broken to
// 1) i==j, matrix[i][j] = true
// 2) the length from i to j is 2 or 3, then, check s[i] == s[j]
// 3) the length from i to j > 3, then, check s[i]==s[j] && matrix[i+1][j-1]
if ( i==j || (s[i]==s[j] && (j-i<2 || matrix[i+1][j-1]) ) ) {
matrix[i][j] = true;
if (longest.size() < j-i+1){
longest = s.substr(i, j-i+1);
}
}
}
}
return longest;
}
// Optimized DP soltuion can be accepted by LeetCode.
string longestPalindrome_dp_opt_way(string s) {
int n = s.size();
if (n<=1) return s;
//Construct a matrix, and consdier matrix[j][i] as s[i] -> s[j] is Palindrome or not.
// ------^^^^^^
// NOTE: it's [j][i] not [i][j]
//Using vector could cause the `Time Limit Error`
//So, use the native array
bool **matrix = (bool**)malloc(n*sizeof(bool*));
int start=0, len=0;
// Dynamic Programming
// 1) if i == j, then matrix[i][j] = true;
// 2) if i != j, then matrix[i][j] = (s[i]==s[j] && matrix[i-1][j+1])
for (int i=0; i<n; i++){
matrix[i] = (bool*)malloc((i+1)*sizeof(bool));
memset(matrix[i], false, (i+1)*sizeof(bool));
matrix[i][i]=true;
for (int j=0; j<=i; j++){
// The following if statement can be broken to
// 1) j==i, matrix[i][j] = true
// 2) the length from j to i is 2 or 3, then, check s[i] == s[j]
// 3) the length from j to i > 3, then, check s[i]==s[j] && matrix[i-1][j+1]
if ( i==j || (s[j]==s[i] && (i-j<2 || matrix[i-1][j+1]) ) ) {
matrix[i][j] = true;
if (len < i-j+1){
start = j;
len = i-j+1;
}
}
}
}
for (int i=0; i<n; i++) {
free (matrix[i]);
}
free(matrix);
return s.substr(start, len);
}
string longestPalindrome(string s) {
return longestPalindrome_dp_way(s);
return longestPalindrome_dp_opt_way(s);
return longestPalindrome_recursive_way2(s);
return longestPalindrome_recursive_way(s);
}
int main(int argc, char**argv)
{
string s = "abacdfgdcaba";
if (argc > 1){
s = argv[1];
}
cout << s << " : " << longestPalindrome(s) << endl;
s = "321012321001232100123210012321001232100123210012321001232100123210012321001232100123210012321001232100123210012321001232100123210012321001232100123210012321001232100123210012321001232100123210012321001232100123210012321001232100123210012321001232100123210012321001232100123210012321001232100123210012321001232100123210012321001232100123210012321001232100123210012321001232100123210012321001232100123210012321001232100123210012321001232100123210012321001232100123210012321001232100123210012321001232100123210012321001232100123210012321001232100123210012321001232100123210012321001232100123210012321001232100123210012321001232100123210012321001232100123210012321001232100123210012321001232100123210012321001232100123210012321001232100123210012321001232100123210012321001232100123210012321001232100123210012321001232100123210012321001232100123210012321001232100123210012321001232100123210012321001232100123210012321001232100123210012321001232100123210012321001232100123210123210012321001232100123210123";
cout << s << " : " << longestPalindrome(s) << endl;
//"illi"
s = "iptmykvjanwiihepqhzupneckpzomgvzmyoybzfynybpfybngttozprjbupciuinpzryritfmyxyppxigitnemanreexcpwscvcwddnfjswgprabdggbgcillisyoskdodzlpbltefiz";
cout << s << " : " << longestPalindrome(s) << endl;
return 0;
}
