README.md

## Leetcode Blog Version
Here is the [Leetcode Blog Version](http://leetcode.com/2011/11/longest-palindromic-substring-part-i.html).
This is a classic problem. I write my first leetcode accepted version based on that.
## My own words version
Brute force + isPalindromic will run out of time.
### Recursive
* `length 1 substring`: all are palindromic
* `length 2 substring`: only if two char are same
What if `length 3`
```
length 3 = length 1 + two char
0 1 2 3 4
a b c b a
+ ^
| |
+---+
0 1 2 4 4
a b c b a
+ ^
| |
+---+
```
* `length n` = inner `length n - 2` is palindromic AND (first char == last char)
### Store `length n` into `P[lenght n][start index]`
* `P[1][3] = true` means that the substring starts from index 3 and 1 char long is palindromic.
* `P[5][2] = true` means that the substring starts from index 2 and 5 char long is NOT palindromic.
a matrix for `abcba`
```
0 1 2 3 4
a b c b a
1 1 1 1 1 1
2 0 0 0 0 -
3 0 1 0 - -
4 0 0 - - -
5 1 - - - -
^
l
e
n
```
With the matrix, we can fill up the martix by `P[len][i] = P[len -2][i + 1] && S[i] == S[i + len -1]`.
Note: the `length 0` is useless, so just using `P[len - 1]` for `len`.