标签：style class blog code java http
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.
Analysis: 想到了Naive的方法，时间复杂度为O(N^3)，知道肯定不是最优的方法，一时也想不出别的方法，于是看了网上的做法，最后采纳了这个人的方法，感觉他讲的比较全面：http://www.programcreek.com/2013/12/leetcode-solution-of-longest-palindromic-substring-java/
1. Naive Approach: The time complexity is O(n^3). If this is submitted to LeetCode onlinejudge, "Time Limit Exceeded".
2. Dynamic Programming:
Let s be the input string, i and j are two indices of the string.
Define a 2-dimension array "table" and let table[i][j] denote whether substring from i to j is palindrome.
Start condition:
table[i][i] == 1; table[i][i+1] == 1 => s.charAt(i) == s.charAt(i+1)
Changing condition:
table[i][j] == 1 => table[i+1][j-1] == 1 && s.charAt(i) == s.charAt(j)
Time O(n^2) Space O(n^2)
3. Smart Algorithm: Time O(n^2), Space O(1)
1 public class Solution { 2 public String longestPalindrome(String s) { 3 if (s == null || s.length() == 0) return null; 4 if (s.length() == 1) return s; 5 String longest = ""; 6 for (int i = 0; i < s.length(); i++) { 7 String temp = helper(s, i, i); 8 if (temp.length() > longest.length()) { 9 longest = temp; 10 } 11 temp = helper(s, i, i + 1); 12 if (temp.length() > longest.length()) { 13 longest = temp; 14 } 15 } 16 return longest; 17 } 18 19 public String helper(String s, int start, int end) { 20 while (start >=0 && end < s.length() && s.charAt(start) == s.charAt(end)) { 21 start--; 22 end++; 23 } 24 start++; 25 end--; 26 return s.substring(start, end + 1); 27 } 28 }
Leetcode: Longest Palindromic Substring,布布扣,bubuko.com
Leetcode: Longest Palindromic Substring
标签：style class blog code java http
原文地址：http://www.cnblogs.com/EdwardLiu/p/3792493.html
