5. Longest Palindromic Substring

class Solution {
    public String longestPalindrome(String s) {
        String ans = "";
        int st = 0, en = 0;
        
        while (st <= s.length() && en <= s.length()) {
            if (isPalindrome(s, st, en)) {
                ans = en - st + 1 > ans.length() ? s.substring(st, en + 1) : ans;
                en++;
            } else {
                st++;
            }
        }
        
        return ans; 
    }
    
    public boolean isPalindrome(String s, int st, int en) {
        if (st<0 || en>=s.length()) return false;
        while (st <= en)
            if (s.charAt(st++) != s.charAt(en--))
                return false;
        return true;
    }
}

힝구..
나중에 나올수도 있다는 점 간과

public String longestPalindrome(String s) {
    if (s == null || s.length() < 1) return "";
    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;
}

Runtime: 23 ms, faster than 90.60% of Java online submissions for Longest Palindromic Substring.
Memory Usage: 39.1 MB, less than 76.48% of Java online submissions for Longest Palindromic Substring.

매 중간을 확인해 주는 방식