[leetcode/lintcode 题解] 最长回文子串 · Longest Palindromic Substring

[hakunamatata11]

[题目描述] 给出一个字符串（假设长度最长为 1000 ），求出它的最长回文子串，你可以假定只有一个满足条件的最长回文串。

在线评测地址: https://www.lintcode.com/problem/longest-palindromic-substring/?utm_source=sc-v2ex-fks0521

[样例] 样例 1:

输入:"abcdzdcab"
输出:"cdzdc"

样例 2:

输入:"aba"
输出:"aba"

[题解]

解法一： 基于中心点枚举的算法，时间复杂度 O(n^2)

public class Solution {
    public String longestPalindrome(String s) {
        if (s == null || s.length() == 0) {
            return "";
        }
        
        int start = 0, len = 0, longest = 0;
        for (int i = 0; i < s.length(); i++) {
            len = findLongestPalindromeFrom(s, i, i);
            if (len > longest) {
                longest = len;
                start = i - len / 2;
            }
            
            len = findLongestPalindromeFrom(s, i, i + 1);
            if (len > longest) {
                longest = len;
                start = i - len / 2 + 1;
            }
        }
        
        return s.substring(start, start + longest);
    }
    
    private int findLongestPalindromeFrom(String s, int left, int right) {
        int len = 0;
        while (left >= 0 && right < s.length()) {
            if (s.charAt(left) != s.charAt(right)) {
                break;
            }
            len += left == right ? 1 : 2;
            left--;
            right++;
        }
        
        return len;
    }
}

解法二： 使用 Manancher's Algorithm，可以在 O(n) 的时间内解决问题

public class Solution {
    public String longestPalindrome(String s) {
        if (s == null || s.length() == 0) {
            return "";
        }
        
        // abc => #a#b#c#
        String str = generateString(s);
        
        int[] palindrome = new int[str.length()];
        int mid = 0, longest = 1;
        palindrome[0] = 1;
        for (int i = 1; i < str.length(); i++) {
            int len = 1; 
            if (mid + longest > i) {
                int mirrorOfI = mid - (i - mid);
                len = Math.min(palindrome[mirrorOfI], mid + longest - i);
            }
            
            while (i + len < str.length() && i - len >= 0) {
                if (str.charAt(i - len) != str.charAt(i + len)) {
                    break;
                }
                len++;
            }
            
            if (len > longest) {
                longest = len;
                mid = i;
            }
            
            palindrome[i] = len;
        }
        
        longest = longest - 1; // remove the extra #
        int start = (mid - 1) / 2 - (longest - 1) / 2;
        return s.substring(start, start + longest);
    }
    
    private String generateString(String s) {
        StringBuilder sb = new StringBuilder();
        for (int i = 0; i < s.length(); i++) {
            sb.append('#');
            sb.append(s.charAt(i));
        }
        sb.append('#');
        
        return sb.toString();
    }
}

[更多解法可参考] https://www.jiuzhang.com/solution/longest-palindromic-substring/?utm_source=sc-v2ex-fks0521