leetcode -- Longest Palindromic Substring

Given a string S, find the longest palindromic substring in S. You may assume that the maximum length of S is 1000, and there exists one unique longest palindromic substring.

O(n*n)。对于每一个字符，以之作为中间元素往左右寻找。注意处理奇偶两种模式：
1. aba
2. abba

public class Solution {
public String longestPalindrome(String s) {
// Start typing your Java solution below
// DO NOT write main() function
String longestPalindrome = "";
for(int i = 0; i < s.length(); i++){

// 需考虑两种情形
// aba 和 abba
String oddPattern = getOddPalindromeSubstring(i, s);
String evenPattern = getEvenPalindromeSubstring(i, s);
if(evenPattern.length() > oddPattern.length()){
if(evenPattern.length() > longestPalindrome.length())
longestPalindrome = evenPattern;
} else {
if(oddPattern.length() > longestPalindrome.length())
longestPalindrome = oddPattern;
}
}

return longestPalindrome;
}

// abba
public String getEvenPalindromeSubstring(int index, String s){
char currentChar = s.charAt(index);
int leftPointer = index - 1;
int rightPointer = index + 1;
//String result = "";
// 回文子串至少包含当前字符
String result = "";
result += currentChar;

/*
if(leftPointer >= 0 && s.charAt(leftPointer) == currentChar){
result = s.substring(leftPointer, index + 1);
leftPointer --;
}
*/

if(rightPointer < s.length() && s.charAt(rightPointer) == currentChar){
result += s.charAt(rightPointer);
rightPointer ++;
}

while(leftPointer >= 0 && rightPointer < s.length()){
if(s.charAt(leftPointer) == s.charAt(rightPointer)){
result = s.substring(leftPointer, rightPointer + 1);
leftPointer --;
rightPointer ++;
continue;
}
break;
}
return result;

}

// aba
public String getOddPalindromeSubstring(int index, String s){
char currentChar = s.charAt(index);
int leftPointer = index - 1;
int rightPointer = index + 1;
//String result = "";
String result = "";
// 回文子串至少包含当前字符
result += currentChar;

while(leftPointer >= 0 && rightPointer < s.length()){
if(s.charAt(leftPointer) == s.charAt(rightPointer)){
result = s.substring(leftPointer, rightPointer + 1);
leftPointer --;
rightPointer ++;
continue;
}
break;
}
return result;
}
}

find a more elegant code：http://leetcode.com/2011/11/longest-palindromic-substring-part-i.html

string expandAroundCenter(string s, int c1, int c2) {
int l = c1, r = c2;
int n = s.length();
while (l >= 0 && r <= n-1 && s[l] == s[r]) {
l--;
r++;
}
return s.substr(l+1, r-l-1);
}

string longestPalindromeSimple(string s) {
int n = s.length();
if (n == 0) return "";
string longest = s.substr(0, 1);  // a single char itself is a palindrome
for (int i = 0; i < n-1; i++) {
string p1 = expandAroundCenter(s, i, i);
if (p1.length() > longest.length())
longest = p1;

string p2 = expandAroundCenter(s, i, i+1);
if (p2.length() > longest.length())
longest = p2;
}
return longest;
}

时间复杂度为O(n)的算法

http://leetcode.com/2011/11/longest-palindromic-substring-part-ii.html