[leetcode][5] Longest Palindromic Substring

5. 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.

Example 1:

Input: "babad"
Output: "bab"
Note: "aba" is also a valid answer.

Example 2:

Input: "cbbd"
Output: "bb"

解析

题目很好理解给定字符串找出最长回文子字符串，首先要清楚怎么样判断是否是回文字符串，然后考虑如何提高效率。

参考答案

自己的（效率低）：

class Solution {
    public String longestPalindrome(String s) {
        String res = "";
        int preStart = 0;
        int preEnd = 0;
        for (int i= 0; i< s.length(); i++) {
            for (int j= i; j < s.length(); j++) {
                if (j-i+1 > res.length() && isPalindrome(s, i, j, preStart, preEnd)) {
                    preStart = i;
                    preEnd = j - 1;
                    res = s.substring(i, j+1);
                }
            }
        }
        return res;
    }
    
    public boolean isPalindrome(String s, int start, int end, int preStart, int preEnd) {
        while (start < end) {
            if (start == preStart && end == preEnd) {
                return true;
            }
            if (s.charAt(start) != s.charAt(end)) {
                return false;
            }
            start++;
            end--;
        }
        
        return true;
    }
}

虽然知道要复用上次的回文数的结果，但是不知道怎么利用，很僵硬。。。虽然通过了，用了163ms。

别人的：

public class Solution {
private int lo, maxLen;

public String longestPalindrome(String s) {
    int len = s.length();
    if (len < 2)
        return s;
    
    for (int i = 0; i < len-1; i++) {
        extendPalindrome(s, i, i);  //assume odd length, try to extend Palindrome as possible
        extendPalindrome(s, i, i+1); //assume even length.
    }
    return s.substring(lo, lo + maxLen);
}

private void extendPalindrome(String s, int j, int k) {
    while (j >= 0 && k < s.length() && s.charAt(j) == s.charAt(k)) {
        j--;
        k++;
    }
    if (maxLen < k - j - 1) {
        lo = j + 1;
        maxLen = k - j - 1;
    }
}}

其实可以从中间开始的，向两边遍历，这样判断回文的效率高很多。用dp应该也可以做，回头再看看。