LeetCode——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, and there exists one unique longest palindromic substring.

给定一个字符串S，找出当中的最长回文字符子串。

1.枚举全部子串，并推断是否是回文串同一时候记录最大长度。超时。

//找出全部子串，并推断是否是回文串，每次记录长度
public static String longestPalindrome1(String s) {
String ret = "";
int maxlen = 0;
for(int i=0;i<s.length();i++){
for(int j=i+1;j<s.length();j++){
String temp = s.substring(i,j+1);
int len = j - i;
if(isPalindrome(temp)){
if(len > maxlen){
ret = temp;
maxlen = len;
}
}
}
}
return ret;
}
public static boolean isPalindrome(String s){
for(int i=0;i<s.length();i++){
if(s.charAt(i) != s.charAt(s.length() - i - 1))
return false;
}
return true;
}

2.由某个中心向两側扩展寻找，找到能够扩展的最大长度。

public static String longestPalindrome(String s) {
if (s.length() <= 1)
return s;String longest = s.substring(0, 1);
for (int i = 0; i < s.length(); i++) {
String tmp = helper(s, i, i);
if (tmp.length() > longest.length()) {
longest = tmp;
}tmp = helper(s, i, i + 1);
if (tmp.length() > longest.length()) {
longest = tmp;
}
}
return longest;
}public static String helper(String s, int begin, int end) {
while (begin >= 0 && end <= s.length() - 1 && s.charAt(begin) == s.charAt(end)) {
begin--;
end++;
}
return s.substring(begin + 1, end);
}