Category: Dynamic Programming
topic
Given two strings A and B, now we need to make A into B through some operations
, the possible operations are:
删除–将字符串 A
delete a character in . Insert – Insert a certain character at a certain position
in string A. replace – replace string A
中的某个字符替换为另一个字符。
Now please ask, change A
into B
How many operations are required at least.
input format
The first line contains the integer n
, representing the string A
length.
The second line contains a string A of length n
。
The third line contains the integer m
, representing the string B
length.
The fourth line contains a string B of length m
。
Strings contain only uppercase letters.
output format
Output an integer representing the minimum number of operations.
data range
1≤n,m≤1000
Input sample:
10
AGTCTGACGC
11
AGTAAGTAGGC
Sample output:
4
java code
import java.util.Scanner;
public class Main {
static Scanner sc = new Scanner(System.in);
static int N = 1010;
static int f[][] = new int[N][N];
static char a[] = new char[N];
static char b[] = new char[N];
public static void main(String[] args) {
int n = sc.nextInt();
sc.nextLine();
char arr[] = sc.nextLine().toCharArray();
int m = sc.nextInt();
sc.nextLine();
char brr[] = sc.nextLine().toCharArray();
for (int i = 0; i <= m; i++) f[0][i] = i;
for (int i = 0; i <= n; i++) f[i][0] = i;
for (int i = 1; i <= n; i++) {
for (int j = 1; j <= m; j++) {
f[i][j] = Math.min(f[i-1][j] + 1, f[i][j-1] + 1);
if (arr[i-1] == brr[j-1]) f[i][j] = Math.min(f[i][j], f[i-1][j-1]);
else f[i][j] = Math.min(f[i][j], f[i-1][j-1] + 1);
}
}
System.out.println(f[n][m]);
}
}