POJ 1159

Description
A palindrome is a symmetrical string, that is, a string read identically from left to right as well as from right to left. You are to write a program which, given a string, determines the minimal number of characters to be inserted into the string in order to obtain a palindrome.

As an example, by inserting 2 characters, the string “Ab3bd” can be transformed into a palindrome (“dAb3bAd” or “Adb3bdA”). However, inserting fewer than 2 characters does not produce a palindrome.

Input
Your program is to read from standard input. The first line contains one integer: the length of the input string N, 3 <= N <= 5000. The second line contains one string with length N. The string is formed from uppercase letters from ‘A’ to ‘Z’, lowercase letters from ‘a’ to ‘z’ and digits from ‘0’ to ‘9’. Uppercase and lowercase letters are to be considered distinct.

Output
Your program is to write to standard output. The first line contains one integer, which is the desired minimal number.

Sample Input

5
Ab3bd

Sample Output

2

题意：
给出一串字符串字母，问最少添加多少字母能把这个字符串变成回文串，相信对于回文串，大家都了解，所以就不赘述了*

思路：
也是看了大神的博客才做出来的，最主要的就是最少次数 = 字符串长度 - 字符串倒叙与 字符串正序的最长公共字串长度*

代码：

import java.util.Scanner;

public class Main{
    private static int num = 0;
    private static char[] array_a = new char[5005];
    private static char[] array_b = new char[5005];
    private static int[][] cc = new int[2][5005];
    public static void main(String[] args) {
    Scanner scanner = new Scanner(System.in);
    num = scanner.nextInt();
    String inputLine = scanner.next();
    array_a = inputLine.toCharArray();
    int data = 0;
    for (int i = inputLine.length() - 1; i >= 0; i--) {
        array_b[data++] = array_a[i];
        cc[0][i] = 0;
    }
    for (int j = 0; j < num; j++) {
        for (int i = 0; i < num; i++) {
        if (array_a[i] == array_b[j]) {
            cc[1][i + 1] = cc[0][i] + 1;
        } else {
            cc[1][i + 1] = cc[1][i] > cc[0][i + 1] ? cc[1][i]:cc[0][i + 1];
        }
        }
        for (int i = 1; i <= num; i++) {
        cc[0][i] = cc[1][i];
        }
    }
    System.out.println(num - cc[1][num]);
    scanner.close();
    }
}