最长回文子串

遍历字符串，逐个判断每个字符，向两边扩散，判断以当前字符为中心，最长回文大小。 

/*** ①中心扩散法* 向左 向右 向左右* ②动态规划优化* 空间换时间*/
class Solution {public static void main(String[] args) {System.out.println(longestPalindrome(new String("a")));}public static String longestPalindrome(String s) {if (s.length() < 2) return s;int l = 0, r = 0;int maxv = 1;for (int i = 0; i < s.length(); i++) {int left = i - 1, right = i + 1;int len = 1;//向左while (left >= 0 && s.charAt(left) == s.charAt(i)) {left--;len++;}//向右while (right < s.length() && s.charAt(right) == s.charAt(i)) {right++;len++;}//向左右while (right < s.length() && left >= 0 && s.charAt(left) == s.charAt(right)) {left--;right++;len += 2;}if (len > maxv) {maxv = len;l = left;r = right;}}return s.substring(l + 1, l + maxv + 1);}
}

动态规划的做法就是：记录已经遍历过的字符串，避免重复判断 

/*** ①中心扩散法* 向左 向右 向左右* ②动态规划优化* 空间换时间*/
class Solution {public static String longestPalindrome(String s) {if (s.length() < 2) return s;int maxv = 1;int ll = 0, rr = 0;boolean dp[][] = new boolean[s.length()][s.length()];for (int r = 1; r < s.length(); r++) {for (int l = 0; l < r; l++) {if (s.charAt(l) == s.charAt(r) && (r - l <= 2 || dp[l + 1][r - 1])) {dp[l][r] = true;if (r - l + 1 > maxv) {maxv = r - l + 1;ll = l;rr = r;}}}}return s.substring(ll, rr + 1);}
}