Given a string s, find the longest palindromic substring in s. You may assume that the maximum length of s is 1000.
Example 1:
Input: "babad" Output: "bab"
Note: "aba" is also a valid answer.
Example 2:
Input: "cbbd" Output: "bb"
Solution 1 in C++
#include <iostream> #include <algorithm> using namespace std; class Solution { public: string longestPalindrome(string s) { int str_len = s.length(); string myPalindrome; string ret; int currentPalindromeLen; int str_remaining=str_len; for(int i=0; i<str_len; i++) { myPalindrome=""; currentPalindromeLen=0; for(int j=i; j<str_len; j++) { myPalindrome += s[j]; currentPalindromeLen++; int idx1=0, idx2=currentPalindromeLen-1; while(idx1<idx2) { if (myPalindrome[idx1]!=myPalindrome[idx2]) break; idx1++; idx2--; } if (idx1>=idx2) { if (myPalindrome.length()>=ret.length()) ret = myPalindrome; } } str_remaining--; if (str_remaining<ret.length()) break; } return ret; } }; int main(void) { Solution s; string testcase1 = "babad"; // the answer should be "bab" or "aba" cout << testcase1 << " -> " << s.longestPalindrome(testcase1) << endl; string testcase2 = "cbbd"; // the answer should be "bb" cout << testcase2 << " -> " << s.longestPalindrome(testcase2) << endl; }
Solution 2 in Java - Time Complexity = O(n^2)
public String longestPalindrome(String s) { if (s == null || s.length() < 1) return ""; int start = 0, end = 0; for (int i = 0; i < s.length(); i++) { int len1 = expandAroundCenter(s, i, i); int len2 = expandAroundCenter(s, i, i + 1); int len = Math.max(len1, len2); if (len > end - start) { start = i - (len - 1) / 2; end = i + len / 2; } } return s.substring(start, end + 1); } private int expandAroundCenter(String s, int left, int right) { int L = left, R = right; while (L >= 0 && R < s.length() && s.charAt(L) == s.charAt(R)) { L--; R++; } return R - L - 1; }