标签:segment imu next tps == put har new ring
Given a string s, find the longest palindromic substring in s. You may assume that the maximum length of s is 1000.
Example:
Input: "babad" Output: "bab" Note: "aba" is also a valid answer.
Example:
Input: "cbbd" Output: "bb"
Similar Questions: Shortest Palindrome Palindrome Permutation Palindrome Pairs Longest Palindromic Subsequence Palindromic Substrings
Next challenges:
思路:对原字符串运行"马拉车"算法(Manacher‘s Algorithm),找到最长回文子串的中心位置所在,然后向两边扩充还原。
关于Manacher‘s Algorithm的学习资料:
https://segmentfault.com/a/1190000003914228
http://www.cnblogs.com/grandyang/p/4475985.html
代码:
1 public class Solution { 2 public String longestPalindrome(String s) { 3 int maxRight = 0, pos = 0, maxLen = 0, maxPos = 0; 4 String rs = "#"; 5 for(int i = 0; i < s.length(); i++) rs = rs + s.charAt(i) + "#"; 6 int[] RL = new int[rs.length()]; 7 for(int i = 0; i < rs.length(); i++) { 8 if(maxRight > i) RL[i] = Math.min(maxRight - i, RL[2 * pos - i]); 9 while(i - RL[i] - 1>= 0 && i + RL[i] + 1< rs.length() && rs.charAt(i - RL[i] - 1) == rs.charAt(i + RL[i] + 1)) RL[i]++; 10 if(RL[i] + i > maxRight) { 11 pos = i; 12 maxRight = RL[i] + i; 13 } 14 if(maxLen < 2 * RL[i] + 1) { 15 maxLen = 2 * RL[i] + 1; 16 maxPos = i; 17 } 18 } 19 return rs.substring(maxPos - RL[maxPos], maxPos + RL[maxPos] + 1).replace("#",""); 20 } 21 }
Leetcode 5. Longest Palindromic Substring
标签:segment imu next tps == put har new ring
原文地址:http://www.cnblogs.com/Deribs4/p/7233601.html