[LeetCode] 5. Longest Palindromic Substring

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本文探讨了如何在给定字符串中找到最长的回文子串，并提供了两种算法实现：一种是通过暴力方法查找所有子串并验证其回文性质；另一种是优化后的算法，通过扩展中心法来确定最长回文子串。

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"

一上来我就想暴力过。。结果94个点只过了82个，算法自以为是O(n^2)实际上是O(n^3)；

class Solution {
    public String longestPalindrome(String s) {
        int length = s.length();
        int max = 1, j;
        String string, result = s.substring(0,1);
        for (int i = 0; i < length; i++){
            j = length-1;
            while (i < j){
                while (s.charAt(j) != s.charAt(i)) j--;
                string = s.substring(i,j+1);
                if (palindromic(string) && max < string.length()){
                    max  = string.length();
                    result = string;
                }
                j--;
            }
        }
        return result;
    }

    private boolean palindromic(String string) {
        int length = string.length();
        int j;
        for (int i = 0; i < length; i++){
            j = length - i - 1;
            if (string.charAt(i) != string.charAt(j)) return false;
        }
        return true;
    }
}

答案算法O(n^2)

public String longestPalindrome(String s) {
    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;
}

很好理解不多说，我的思路是取一个字序列出来，再判断它是否为回文序列；

答案的思路是求以i为中心的最长回文子序列，回文子序列有两种“aba”和“abba”，因此有len1和len2；

substring（1，10）是从第一位截到第九位，第九位截进去而第十位不截

github