lc5 最长回文子串

1 动态规划

class Solution {
    public String longestPalindrome(String s) {
        int len = s.length();
        if (len < 2) {
            return s;
        }
        int maxLen = 1, begin = 0;
        boolean[][] dp = new boolean[len][len];
        for (int i = 0; i < len; i++){
            dp[i][i] = true;
        }
        char[] chars = s.toCharArray();
        for (int L = 2; L <= len; L++) {
            for (int i = 0; i < len; i++) {
                int j = L + i - 1;
                if (j >= len) {
                    break;
                }
                if (chars[i] != chars[j]) {
                    dp[i][j] = false;
                } else {
                    if (j - i < 3) {
                        dp[i][j] = true;
                    } else {
                        dp[i][j] = dp[i+1][j-1];
                    }
                }
                if (dp[i][j] && j - i + 1 > maxLen) {
                    maxLen = j - i + 1;
                    begin = i;
                }
            }
        }
        return s.substring(begin, begin + maxLen);
    
    }
}

2 中心扩展算法

class Solution {
    public String longestPalindrome(String s) {
        int length = s.length();
        if (length < 2) {
            return s;
        }
        int start = 0, end = 0;
        for (int i = 0; i < length; i++) {
            int len1 = expand(s, i, i);
            int len2 = expand(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);
    }
    public int expand(String s, int left, int right) {
        while (left >= 0 && right < s.length() && s.charAt(left) == s.charAt(right)) {
            --left;
            ++right;
        }
        return right - left - 1; 
    }
}

3 Manacher

略

lc

标签：head,ListNode,lc,int,0921,next,return,curr,回文
From： https://www.cnblogs.com/beichuang/p/16717289.html