# 问题描述

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. 所谓回文字符串，就是一个字符串，从左到右读和从右到左读是完全一样的。比如”a” , “aaabbaaa”

# 解决方案

## 1.暴力方案(Brute Force)

#include "stdafx.h"
#include <iostream>
#include <string>
using namespace std;
char result={0};

bool isHuiwen(int begin,int end,char* s)
{
if (end==begin||end<begin)
{
return true;
}
if (s[begin]!=s[end])
{
return false;
}
return isHuiwen(begin+1,end-1,s);
}

char* longestHuiwen(int length,char* s)
{
int begin = 0,end=0,sum=0;
for (int i=0;i<length;i++)
{
for (int j=0;j<=i;j++)
{
if (isHuiwen(j,i,s))
{
if (i-j>=sum)
{
sum = i -j;
begin = j;
end = i;
}

}

}
}
strncpy(result,s+begin,sum+1);//由0开始计数
return result;
}

int _tmain(int argc, _TCHAR* argv[])
{
char* s = "abcabaaaabbacabbaa";
char* r_s = longestHuiwen(18,s);
return 0;
}

## 2.问题转换为求最长相似子串

Approach #1 (Longest Common Substring) [Accepted]

Common mistake

Some people will be tempted to come up with a quick solution, which is unfortunately flawed (however can be corrected easily):

Reverse S and become S′. Find the longest common substring between S and S​′, which must also be the longest palindromic substring.This seemed to work, let’s see some examples below.

For example, S=”caba” S′=”abac”

The longest common substring between S and S​′ is ”aba”, which is the answer. Let’s try another example: S=”abacdfgdcaba” S′=”abacdgfdcaba”

The longest common substring between S and S​′ is ”abacd” Clearly, this is not a valid palindrome.

# 其他三种解法

## Approach #3 (Dynamic Programming) [Accepted]

To improve over the brute force solution, we first observe how we can avoid unnecessary re-computation while validating palindromes. Consider the case ”ababa” ”ababa”. If we already knew that ”bab” ”bab” is a palindrome, it is obvious that ”ababa” ”ababa” must be a palindrome since the two left and right end letters are the same.

We define P(i,j)P(i,j) as following:

P(i,j)={true, if the substring Si…Sj is a palindrome false, otherwise.

P(i,j)={true,if the substring Si…Sj is a palindromefalse,otherwise. Therefore,

P(i, j) = ( P(i+1, j-1) \text{ and } S_i == S_j ) P(i,j)=(P(i+1,j−1) and S​i==S​j)

The base cases are:

P(i, i) = true P(i,i)=true

P(i, i+1) = ( S_i == S_{i+1} ) P(i,i+1)=(S​i ==Si+1)

This yields a straight forward DP solution, which we first initialize the one and two letters palindromes, and work our way up finding all three letters palindromes, and so on…

Complexity Analysis

Time complexity : O(n^2)O(n​2). This gives us a runtime complexity of O(n^2)O(n2).

Space complexity : O(n^2)O(n​2). It uses O(n^2)O(n2) space to store the table.

Could you improve the above space complexity further and how?

## Approach #4 (Expand Around Center) [Accepted]

In fact, we could solve it in O(n^2)O(n​2 ) time using only constant space.

We observe that a palindrome mirrors around its center. Therefore, a palindrome can be expanded from its center, and there are only 2n - 12n−1 such centers.

You might be asking why there are 2n - 12n−1 but not nn centers? The reason is the center of a palindrome can be in between two letters. Such palindromes have even number of letters (such as ”abba””abba”) and its center are between the two ‘b”b’s.

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; } Complexity Analysis

Time complexity : O(n^2)O(n​2​​ ). Since expanding a palindrome around its center could take O(n)O(n) time, the overall complexity is O(n^2)O(n​2​​ ).

Space complexity : O(1)O(1).

## Approach #5 (Manacher’s Algorithm) [Accepted]

There is even an O(n)O(n) algorithm called Manacher’s algorithm, explained here in detail. However, it is a non-trivial algorithm, and no one expects you to come up with this algorithm in a 45 minutes coding session. But, please go ahead and understand it, I promise it will be a lot of fun.

# 参考代码

c代码

char* longestPalindrome(char* s) {
int i,length=strlen(s);
char* new_s;
new_s=malloc(sizeof(char)*(2*length + 2));
new_s='\$';
new_s='#';

for(i=0;i<length;i++)
{
*(new_s+2*i+2)=s[i];
*(new_s+2*i+3)='#';

}
int len=2*length + 2;
int* r;
r=malloc(sizeof(int)*len);
r=0;
int center=1;
int max_right=0;
for(i=1;i<len;i++)
{
if(i<max_right)
{
if( (max_right-i)> r[2*center-i] )
r[i]=r[2*center-i];
else
r[i]=(max_right-i);
}
else r[i]=1;
while(new_s[i-r[i]]==new_s[i+r[i]] && i-r[i]>0 && i+r[i]<len)
{
r[i]++;
}

if(i+r[i] > max_right)
{
center = i;
max_right = i+r[i];

}

}
int max_r = 0;
int j=0;
for(i=1;i<len;i++)
{
if( max_r<r[i])
{
j=i;
max_r= r[i];
}
}
int m=(j-(max_r-2)-2)/2;
int n=(j+(max_r-2)-2)/2;
char *c;
c=malloc((max_r)*sizeof(char));

int x=0;
for(i=m;i<=n,x<max_r-1;i++)
{
c[x]=s[i];
x++;
}
*(c+max_r-1)='\0';
return c;
free(r);
free(new_s);
free(c);

}

c++代码

string longestPalindrome(string s) {
if (s.empty()) return"";
if (s.size() == 1) return s;
int min_start = 0, max_len = 1;
for (int i = 0; i < s.size();) {
if (s.size() - i <= max_len / 2) break;
int j = i, k = i;
while (k < s.size()-1 && s[k+1] == s[k]) ++k; // Skip duplicate characters.
i = k+1;
while (k < s.size()-1 && j > 0 && s[k + 1] == s[j - 1]) { ++k; --j; } // Expand.int new_len = k - j + 1;
if (new_len > max_len) { min_start = j; max_len = new_len; }
}
return s.substr(min_start, max_len);
}

python参考代码

def longestPalindrome(self, s):
res = ""
for i in xrange(len(s)):
# odd case, like "aba"
tmp = self.helper(s, i, i)
if len(tmp) > len(res):
res = tmp
# even case, like "abba"
tmp = self.helper(s, i, i+1)
if len(tmp) > len(res):
res = tmp
return res

# get the longest palindrome, l, r are the middle indexes
# from inner to outer
def helper(self, s, l, r):
while l >= 0 and r < len(s) and s[l] == s[r]:
l -= 1; r += 1
return s[l+1:r]

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