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

 1 public class Solution {
 2     public String longestPalindrome(String s) {
 3         // Start typing your Java solution below
 4         // DO NOT write main() function
 5         String longestPalindrome = "";
 6         for(int i = 0; i < s.length(); i++){
 7         
 8             // 需考虑两种情形
 9             // aba 和 abba
10             String oddPattern = getOddPalindromeSubstring(i, s);
11             String evenPattern = getEvenPalindromeSubstring(i, s);
12             if(evenPattern.length() > oddPattern.length()){
13                 if(evenPattern.length() > longestPalindrome.length())
14                     longestPalindrome = evenPattern;
15             } else {
16                 if(oddPattern.length() > longestPalindrome.length())
17                     longestPalindrome = oddPattern;
18             }
19         }
20         
21         return longestPalindrome;
22     }
23     
24     // abba
25     public String getEvenPalindromeSubstring(int index, String s){
26         char currentChar = s.charAt(index);
27         int leftPointer = index - 1;
28         int rightPointer = index + 1;
29         //String result = "";
30         // 回文子串至少包含当前字符
31         String result = "";
32         result += currentChar;
33         
34         /*
35         if(leftPointer >= 0 && s.charAt(leftPointer) == currentChar){
36             result = s.substring(leftPointer, index + 1);
37             leftPointer --;
38         }
39         */
40         
41         if(rightPointer < s.length() && s.charAt(rightPointer) == currentChar){
42             result += s.charAt(rightPointer);
43             rightPointer ++;
44         }
45         
46         while(leftPointer >= 0 && rightPointer < s.length()){
47             if(s.charAt(leftPointer) == s.charAt(rightPointer)){
48                 result = s.substring(leftPointer, rightPointer + 1);
49                 leftPointer --;
50                 rightPointer ++;
51                 continue;
52             }
53             break;
54         }
55         return result;
56         
57     }
58     
59     // aba
60     public String getOddPalindromeSubstring(int index, String s){
61         char currentChar = s.charAt(index);
62         int leftPointer = index - 1;
63         int rightPointer = index + 1;
64         //String result = "";
65         String result = "";
66         // 回文子串至少包含当前字符
67         result += currentChar;
68         
69         while(leftPointer >= 0 && rightPointer < s.length()){
70             if(s.charAt(leftPointer) == s.charAt(rightPointer)){
71                 result = s.substring(leftPointer, rightPointer + 1);
72                 leftPointer --;
73                 rightPointer ++;
74                 continue;
75             }
76             break;
77         }
78         return result;
79     }
80 }

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

 1 string expandAroundCenter(string s, int c1, int c2) {
 2   int l = c1, r = c2;
 3   int n = s.length();
 4   while (l >= 0 && r <= n-1 && s[l] == s[r]) {
 5     l--;
 6     r++;
 7   }
 8   return s.substr(l+1, r-l-1);
 9 }
10  
11 string longestPalindromeSimple(string s) {
12   int n = s.length();
13   if (n == 0) return "";
14   string longest = s.substr(0, 1);  // a single char itself is a palindrome
15   for (int i = 0; i < n-1; i++) {
16     string p1 = expandAroundCenter(s, i, i);
17     if (p1.length() > longest.length())
18       longest = p1;
19  
20     string p2 = expandAroundCenter(s, i, i+1);
21     if (p2.length() > longest.length())
22       longest = p2;
23   }
24   return longest;
25 }

时间复杂度为O(n)的算法

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

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原文地址：https://www.cnblogs.com/feiling/p/3139849.html