杭电1513_Palindrome——java（LCS+滚动数组）

最新推荐文章于 2020-07-09 11:00:48 发布

原创最新推荐文章于 2020-07-09 11:00:48 发布 · 305 阅读

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本文介绍了一种算法，用于确定将任意字符串转换为回文串所需的最少字符插入数量。通过寻找原始字符串及其逆序之间的最长公共子序列，进而得出需要插入的字符数。

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Problem 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

分析：这道题只要将原字符串逆序，与原字符串找出最长公共子序列的长度x，再用原字符串长度n减去x，即为答案。如果数据量的问题，如果数组开大会内存超出，所以利用滚动数组解决（LCS在DP的时候只要上个状态的结果，所以取模运算可以解决）。

import com.sun.javafx.image.IntPixelGetter;

import java.util.*;

public class Main {
    public static long count = 0;

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        while(sc.hasNextInt()){
            int n = sc.nextInt();
            int[][] dp = new int[2][5001];
            int[] a = new int[5001];
            int[] b = new int[5001];
            String temp = sc.nextLine();
            String data = sc.nextLine();
            for(int i = 1;i <= n;i++){
                char c = data.charAt(i-1);
                a[i] = c;
                b[n+1-i] = c;
            }
            for(int i = 1;i <= n;i++){
                for(int j = 1;j <= n;j++){
                    if(a[i] == b[j]){
                        dp[i%2][j] = dp[(i-1)%2][j-1]+1;
                    }
                    else{
                        dp[i%2][j] = max(dp[i%2][j-1],dp[(i-1)%2][j]);
                    }
                }
            }
            System.out.println(n - dp[n%2][n]);

        }
    }
    public static int max(int a, int b){
        if(a > b)
            return a;
        return b;
    }
}