5. Longest Palindromic Substring

题目

给定一个字符串 s，在 s 中找到最长的回文子串，s 的最大长度是 1000。

题目难度：中等

描述

Medium

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"

题解

找字符串中最小回文子串，2 种方法。

一、动态规划(DP)

定义一个数组 dp，d[i][j] 是回文的前提：

1. 第 i 个字符和第 j 个字符相等
2. j-i <= 2 (判断边界：当 i== j时，如 "cbbd" 中，i = 1, j=1, 那么单纯判断 d[i+1][j-1]= d[2][3] 是不符合要求的，所以如何 j-i > 2，则该 i 到 j的子串直接就可以判断不是回文)，在 j-i <= 2时候，才可以判断d[i+1][j-1] 是否是回文

Java

// dp time: O(n^2) space: O(n^2)
public static String longestPalindromeDP(String s) {
  if(s == null || s.length() == 0) return s;
  String res = "";
  int max = 0;
  boolean[][] dp = new boolean[s.length()][s.length()];
  for(int j=0; j max){
          max = j-i+1;
          res = s.substring(i, j+1);
        }
      }
    }
  }
  return res;
}

Python

def LongestPalindromicSubstring_DP(s):
    if not s or len(s) == 0:
        return s
    max = 0
    res = ''
    dp = [[False for x in range(len(s))] for j in range(len(s))]
    for j in range(len(s)):
        for i in range(0, j + 1):
            dp[i][j] = s[i] == s[j] and ((j - i <= 2) or dp[i + 1][j - 1])
            if dp[i][j]:
                if j - i + 1 > max:
                    max = j - i + 1
                    res = s[i: j + 1]
    return res

二、中心扩散

如字符串 “baab”，先以 b 为中心，往前后扩散 1 个，2 个值为起点，如果是回文，再向后扩散

Java

static String res = "";
public static String longestPalindromeCenterSpread(String s) {
  if(s == null || s.length() == 0) return s;
  for(int j=0; j= 0 && right < s.length() && s.charAt(left) == s.charAt(right)){
    left--;
    right++;
  }
  String cur = s.substring(left+1, right);
  // System.out.println(cur);
  if(cur.length() > res.length()){
    res = cur;
  }
}
// 运行结果
a

b
abba
b

a

abba

Python

res = ''
def helper(s, l, r):
    while l >= 0 and r < len(s) and s[l] == s[r]:
        l -= 1
        r += 1
    global res
    cur = s[l + 1:r]
    if len(res) < len(cur):
        res = cur

def LongestPalindromicSubstring(s):
    if not s or len(s) == 0:
        return s

    global res
    for i in range(len(s)):
        helper(s, i, i)
        helper(s, i, i + 1)
    return res