Problem description
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"
Code
class Solution {
int maxLength = 2005;
char[] c = new char[maxLength];
char[] re = new char[maxLength];
int[] res = new int[maxLength];
void change(String s) {
char b = '$';
char m = '#';
int len = s.length();
c[0] = b;
c[1] = m;
for(int i = 0; i < len; i++) {
c[i * 2 + 2] = s.charAt(i);
c[i * 2 + 3] = m;
}
}
String manacher(int len) {
res[0] = 1;
int id = 0;
int mx = 1;
for(int i = 1; i < len; i++) {
if(mx > i) {
res[i] = res[2 * id - i] < mx - i ? res[2 * id - i] : mx - i;
} else {
res[i] = 1;
}
for(; c[i - res[i]] == c[i + res[i]]; res[i]++);
if(mx < i + res[i]) {
mx = i + res[i];
id = i;
}
}
int max = res[0];
int point = 0;
for(int i = 1; i < len; i++) {
if(max < res[i]) {
max = res[i];
point = i;
}
}
int r, l;
l = point - max + 1;
r = point + max - 1;
String ress = "";
int le = 0;
for(int i = l; i <= r; i++) {
if(c[i] != '#') {
ress += c[i];
}
}
return ress;
}
public String longestPalindrome(String s) {
int len = s.length();
change(s);
return manacher(2 * len + 1);
}
}
Analysis
- manacher算法（詳細(xì)分析待補(bǔ)）
