LeetCode 题解（67）: Longest Palindromic Substring
题目：
Given a string S, find the longest palindromic substring in S. You may assume that the maximum length ofS is 1000, and there exists one unique longest palindromic substring.
题解：
首先学习了一下LeetCode讨论区里的DP解法，空间复杂度O(n^2)，Memory Limit Exceeded无法通过：
class Solution {
public:
string longestPalindrome(string s) {
if(!s.length())
return NULL;
int** score = new int*[s.length()];
for(int i = 0; i < s.length(); i++)
score[i] = new int[s.length()];
int max = 1, start = 0, end = 0;
for(int row = 0; row < s.length(); row++)
for(int col = 0; col < s.length(); col++) {
if(row == col)
score[row][col] = 1;
else if(row == col - 1 && s[row] == s[col]) {
score[row][col] = 2;
max = 2;
start = row;
end = col;
}
else
score[row][col] = 0;
}
for(int outer = s.length() - 2; outer > 0; outer--)
for(int inner = 0; inner < outer; inner++) {
int endIndex = inner+s.length()-outer;
if(s[inner] == s[endIndex] && score[inner+1][endIndex-1] > 0) {
score[inner][endIndex] = score[inner+1][endIndex-1] + 1;
if(score[inner][endIndex] > max) {
max = score[inner][endIndex];
start = inner;
end = endIndex;
}
}
}
return s.substr(start, end - start + 1);
}
};
只好继续寻求减少空间的方法，继续学习 LeetCode讨论区的解法：
A simpler approach, O(N2) time and O(1) space:
In fact, we could solve it in O(N2) time without any extra space.
We observe that a palindrome mirrors around its center. Therefore, a palindrome can be expanded from its center, and there are only 2N-1 such centers.
2N-1个中心是怎么得来的？N个字符，每个可以作为一个中心进行扩展，同时，N-1个字符之间的缝隙也可以作为中心扩展（如"abba"，ab和ba中的缝隙向两边扩展得到abba)，所以总共有N+N-1=2N-1个中心。这种向两边expand的方法比较直观，代码也好写，于是得到此150ms解法：
class Solution {
public:
string longestPalindrome(string s) {
if(!s.length())
return NULL;
int max = 1, start = 0, end = 0;
for(int i = 1; i <= s.length(); i++) {
int localMax = 1, localStart = i-1, localEnd = i-1;
for(int j = 0; j < min(i-1, static_cast<int>(s.length()-i)); j++) {
if(s[i-j-2] == s[i+j]) {
localMax += 2;
localStart = i-j-2;
localEnd = i + j;
}
else
break;
}
if(localMax > max) {
max = localMax;
start = localStart;
end = localEnd;
}
}
for(int m = 0; m < s.length()-1; m++) {
if(s[m] == s[m+1]) {
int localMax = 2, localStart = m, localEnd = m + 1;
for(int n = 0; n < min(m, static_cast<int>(s.length()-m-2)); n++) {
if(s[m-n-1] == s[m+n+2]) {
localMax += 2;
localStart = m - n - 1;
localEnd = m + n + 2;
}
else
break;
}
if(localMax > max) {
max = localMax;
start = localStart;
end = localEnd;
}
}
}
return s.substr(start, end-start+1);
}
};
Java版：
尝试把字符和缝隙两种情况写在一起了，逻辑上有点混乱，最后把自己绕迷糊了：
public class Solution {
public String longestPalindrome(String s) {
int max = 1, start = 0, end = 0;
for(int i = 0; i < s.length() * 2 - 1; i++) {
int localMax = 0, localStart = 0, localEnd = 0;
if(i % 2 == 0) {
localMax = 1;
for(int j = 0; j < Math.min(i / 2, s.length() - i / 2 - 1); j++) {
if(s.charAt(i / 2 - j - 1) == s.charAt(i / 2 + j + 1)) {
localMax += 2;
localStart = i / 2 - j - 1;
localEnd = i / 2 + j + 1;
}
else
break;
}
} else {
for(int j = 0; j <= Math.min(i / 2, s.length() - i / 2 - 2); j++) {
if(s.charAt(i / 2 - j) == s.charAt(i / 2 + j + 1)) {
localMax += 2;
localStart = i / 2 - j;
localEnd = i / 2 + j + 1;
}
else
break;
}
}
if(localMax > max) {
max = localMax;
start = localStart;
end = localEnd;
}
}
return s.substring(start, end+1);
}
}
Python版：
class Solution:
# @return a string
def longestPalindrome(self, s):
longest = 1
begin = 0
end = 0
for i in range(0, len(s)):
localLongest = 1
localBegin = 0
localEnd = 0
for j in range(0, min(i, len(s)-i-1)):
if s[i-j-1] == s[i+j+1]:
localLongest += 2
localBegin = i - j - 1
localEnd = i + j + 1
else:
break
if localLongest > longest:
longest = localLongest
begin = localBegin
end = localEnd
for m in range(0, len(s)-1):
if s[m] == s[m+1]:
localLongest = 2
localBegin = m
localEnd = m + 1
for n in range(0, min(m, len(s)-m-2)):
if s[m-n-1] == s[m+n+2]:
localLongest += 2
localBegin = m - n - 1
localEnd = m + n + 2
else:
break
if localLongest > longest:
longest = localLongest
begin = localBegin
end = localEnd
return s[begin:end+1]