思路：动态规划
public String longestPalindrome(String s) {
// 注意：判断输入范围
if (s == null || s.length() == 0) {
return "";
}
// boolean的默认值是false，不用初始化了
boolean[][] dp = new boolean[s.length()][s.length()];
int maxSubLen = 0;
String maxSubStr = null;
for (int len = 0; len < s.length(); len++) {
for (int i = 0; i < s.length() - len; i++) {
int j = i + len;
if (i == j) {
// 单个字符肯定是回文串
dp[i][j] = true;
} else if (j - i == 1) {
// 两个字符如果相等也是回文串
dp[i][j] = s.charAt(i) == s.charAt(j);
} else {
dp[i][j] = (s.charAt(i) == s.charAt(j)) && dp[i + 1][j - 1];
}
if (dp[i][j] && maxSubLen < (j - i + 1)) {
maxSubStr = s.substring(i, j + 1);
}
}
}
return maxSubStr;
}