为什么80%的码农都做不了架构师？>>> 
问题：
Given a string, your task is to count how many palindromic substrings in this string.
The substrings with different start indexes or end indexes are counted as different substrings even they consist of same characters.
Example 1:
Input: "abc" Output: 3 Explanation: Three palindromic strings: "a", "b", "c".
Example 2:
Input: "aaa" Output: 6 Explanation: Six palindromic strings: "a", "a", "a", "aa", "aa", "aaa".
Note:
- The input string length won't exceed 1000.
解决：
① 计算字符串中回文子串的个数，使用动态规划解决：
d[i][j]表示从i到j的字符串为是否回文，真则为（1），否则为假（0），
那么d[i][j]为真的前提是：头尾两个字符串相同并且去掉头尾以后的字串也是回文（即d[i+1][j-1]为真），这里面要注意特殊情况，即：去掉头尾以后为空串，所以如果j-i<3，并且头尾相等，也是回文的。
class Solution { //22ms
public int countSubstrings(String s) {
int len = s.length();
int res = 0;
boolean[][] dp = new boolean[len][len];
for (int i = len - 1;i >= 0;i --){
for (int j = i;j < len;j ++){
dp[i][j] = ((s.charAt(i) == s.charAt(j) && ((j - i < 3) || dp[i + 1][j - 1])));
if (dp[i][j]){
res ++;
}
}
}
return res;
}
}
② 从中间向两边递归判断回文字符串。
class Solution { //12ms
public int countSubstrings(String s) {
if (s == null || s.length() == 0) return 0;
int len = s.length();
int res = 0;
for (int i = 0;i < len;i ++){
res += dfs(s,i,i);
res += dfs(s,i,i + 1);
}
return res;
}
public int dfs(String s,int left,int right){
int res = 0;
while(left >= 0 && right < s.length() && s.charAt(left) == s.charAt(right)){
left --;
right ++;
res ++;
}
return res;
}
}