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添加字符

[编程题]添加字符
  • 热度指数:2831 时间限制:C/C++ 1秒,其他语言2秒 空间限制:C/C++ 32M,其他语言64M
  • 算法知识视频讲解
牛牛手里有一个字符串A,羊羊的手里有一个字符串B,B的长度大于等于A,所以牛牛想把A串变得和B串一样长,这样羊羊就愿意和牛牛一起玩了。
而且A的长度增加到和B串一样长的时候,对应的每一位相等的越多,羊羊就越喜欢。比如"abc"和"abd"对应相等的位数为2,为前两位。
牛牛可以在A的开头或者结尾添加任意字符,使得长度和B一样。现在问牛牛对A串添加完字符之后,不相等的位数最少有多少位?

输入描述:
第一行为字符串A,第二行为字符串B,A的场地小于等于B的长度,B的长度小于等于50.字符均为小写字母。


输出描述:
输出一个整数表示A串添加完字符之后,不相等的位数最少有多少位?
示例1

输入

abe
cabc

输出

1
import java.util.*;
public class Main{
    public static void main(String[] args) {
        Scanner in = new  Scanner(System.in);
        String s1 = in.nextLine();
        String s2 = in.nextLine();
        int maxlen = 0;
        for(int i=0; i<=s2.length()-s1.length(); i++) {
            int count = 0;
            for(int j=0; j<s1.length(); j++) {
                if(s1.charAt(j) == s2.charAt(i+j)) {
                  count++;
                }
            }
            maxlen = Math.max(maxlen, count);
        }
        System.out.println(s1.length() - maxlen);
    }
}

编辑于 2017-05-26 11:14:07 回复(9)
#include<iostream>
#include<string.h>
using namespace std;

int CmpStr(char *strA, char *strB)
{
int cont = 0, max = 81;
int NA = strlen(strA), N = strlen(strB) - strlen(strA);
do
{
for (int i = 0; i < NA; i++)
{
if (strA[i] != strB[i + N])
cont++;
}
if (max > cont)
max = cont;
cont = 0;
} while (N--);
return max;
}

int main()
{
char strA[81], strB[81];
cin >> strA >> strB;
cout << CmpStr(strA, strB) << endl;
return 0;
}
发表于 2017-06-22 12:40:30 回复(0)
就是求两个串的不一样的位数,,就是一个循环。
发表于 2017-10-27 22:03:33 回复(0)
A = input()
B = input()
na,nb = len(A),len(B)
res = []
for i in range(nb-na+1):
    count = 0
    for j in range(na):
        if B[i+j]!=A[j]:
            count += 1
    res.append(count)
print(min(res))

发表于 2020-05-03 18:01:38 回复(1)
var s1 = readline();
var s2 = readline();

if(s2.indexOf(s1) !== -1){
    print(0);
}
else{
    var c = [];
    for(var i = 0; i < s2.length - s1.length + 1; i++){
        var s11 = s2.slice(0, i) + s1 + s2.slice(i + s1.length);
        c.push(sameCount(s11, s2));
    }
    print(Math.min.apply(this, c));
}

function sameCount (a, b) {
    var count = 0;
    for(var i = 0; i < a.length; i++){
        if(a.charAt(i) == b.charAt(i)){
            count++;
        }
    }
    return a.length - count;
}
给JavaScript草存在感
发表于 2017-05-22 16:11:07 回复(2)
哈哈哈,想说一个和答案无关的事情,牛牛一定很喜欢羊羊!他付出了那么多,搞个算法题,只为了让羊羊喜欢他😄😉
发表于 2020-04-09 11:22:45 回复(0)
本质上就是求最大重合字符个数,除了短的字符串内部不能更换,两边的都可以添加成重复字符,所以结果就是len1-max

import java.util.Scanner;

public class Main3 {

    public static void main(String[] args) {
        Scanner scanner=new Scanner(System.in);
        
        String s1=scanner.nextLine();
        String s2=scanner.nextLine();
        
        int len1=s1.length();
        int len2=s2.length();
        int count=0,max=0;
        
        for(int i=0;i<=len2-len1;i++){
            for(int j=0;j<len1;j++){
                if(s1.charAt(j)==s2.charAt(j+i))
                    count++;
            }
            if(max<count){
                max=count;
            }
            
            count=0;
        }
        
        System.out.println(len1-max);
        scanner.close();
    }
}
发表于 2018-05-15 10:46:35 回复(0)
import java.util.Scanner;

/*
 * 直接比较已有字符串中不相等的个数即可
 * */
public class MaxSameLen {
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        while (scanner.hasNext()) {
            String A = scanner.next();
            String B = scanner.next();

            int min = Integer.MAX_VALUE;
//            以A为模式串
            for (int i = 0; i <= B.length() - A.length(); i++) {
                int tmp = 0;
                for (int j = 0; j < A.length(); j++) {
                    if (A.charAt(j) != B.charAt(i + j)) {
                        tmp++;
                    }
                }
//                如果是子串的话,肯定为0
                min = Math.min(min, tmp);
                if (min == 0) {
                    break;
                }
            }
            System.out.println(min);
        }
    }


}
/*
adaabc
aababbc
2



* */
发表于 2018-04-05 17:16:01 回复(0)
 
<?php
$array[0]=str_split(trim(fgets(STDIN)));
$array[1]=str_split(trim(fgets(STDIN)));
$aa= [];
$cha= count($array[1])-count($array[0]);
for($i=0;$i<=$cha;$i++){
    $aa[$i] = 0;
    foreach($array[0] as$k=>$v){
        if($v!= $array[1][$i+$k]) $aa[$i]++;
    }
}
echomin($aa);

发表于 2018-03-22 17:46:00 回复(0)

adaabc

aababbc

为什么输出是2??????????????????????????????

发表于 2017-10-20 04:26:13 回复(0)
#include<iostream>
#include<cstring>
using namespace std;

int main(){
char A[55],B[55];
cin>>A>>B;
int count=strlen(A);
int sum=0,t=0;
for(int i=0,j=0;i<strlen(B),j<strlen(A),t<=strlen(B)-strlen(A);i++,j++){
    if(B[i]!=A[j])  sum++;
    if(j==strlen(A)-1){
        if(sum<count)count=sum;
            i=i-j;
            j=-1;
            t++;
            sum=0;
        }
    }
cout<<count<<endl;
}
编辑于 2017-09-08 16:54:12 回复(0)

python3

def least_different(a,b):
    la = len(a)
    lb = len(b)
    det = lb-la
    count_list = [0 for j in range(det+1)]
    for i in range(la):
        for j in range(det+1):
            if a[i] != b[i+j]:
                count_list[j] +=1
    return min(count_list)

a = input()
b = input()
print(least_different(a,b))
发表于 2017-09-07 16:04:11 回复(0)
import java.util.*;
public class Main {
public static void main(String[] args) {
// TODO Auto-generated method stub
        Scanner sc=new Scanner(System.in);
        while(sc.hasNext()){
        String s1=sc.nextLine();
        String s2=sc.nextLine();
        int length=s2.length()-s1.length();
        int maxCount=0;
        for(int i=0;i<=length;i++){
                int count=0;
        for(int j=0;j<s1.length();j++){
        if(s1.charAt(j)==s2.charAt(i+j)){
        count++;
        maxCount=(maxCount>count)?maxCount:count;
        }
        }
        }
        System.out.println(s1.length()-maxCount);
        }
        sc.close();
}

}

发表于 2017-08-20 11:57:49 回复(0)
import java.util.Scanner;


public class Main3 {

public static void main(String[] args) {
Scanner scanner=new Scanner(System.in);
String A=scanner.nextLine();
String B=scanner.nextLine();
scanner.close();
int m=A.length();
int n=B.length();
int count;
int maxLen=0;
for(int start=0,end=m;end<=n;start++,end++){
//依次从b中取与a等长的子串,判断子串中与a中重复字符个数,找到最大值即可
String BSubstring=B.substring(start, end);
count=0;
for(int i=0;i<m;i++){
if(A.charAt(i)==BSubstring.charAt(i)){
count++;
}
}
if(count>maxLen){
maxLen=count;
}
}
System.out.println(A.length()-maxLen);

}

}

发表于 2017-07-24 21:00:40 回复(0)
package 全国统一模拟第三场;

import java.util.Scanner;

public class Main_3 {
	
	public static void main(String[] args) {
		Scanner sc=new Scanner(System.in);
		while(sc.hasNext())
		{
			String str1=sc.next();
			String str2=sc.next();
			Max_diff(str1,str2);
		}
		sc.close();
	}
	public static void Max_diff(String str1,String str2)
	{
		int min_dif=Integer.MAX_VALUE;
		int diff=0;		
		for(int i=0;i<=str2.length()-str1.length();i++)
		{
			diff=diff_num(str1, str2.substring(i,i+str1.length()));
			min_dif=min_dif<diff?min_dif:diff;			
		}
		System.out.println(min_dif);
	}
	
	public static int  diff_char(String str1,String str2,int diff)
	{
		if(diff==0)
		{						
			return diff_num(str1, str2);
		}
		return Math.min(diff_char(str1, str2.substring(0, str2.length()-1),diff-1), diff_char(str1, str2.substring(1, str2.length()), diff-1));
	}
	public static int diff_num(String str1,String str2)
	{
		int res=0;
		for(int i=0;i<str1.length();i++)
		{
			if(str1.charAt(i)!=str2.charAt(i))
				res++;
		}
		return res;
	}
}


发表于 2017-07-19 22:14:55 回复(0)
def f(a, b):
    a = '0' + a
    b = '0' + b
    la = len(a)
    lb = len(b)
    if la == lb:
        ret = 0
        for i in range(1, la):
            if a[i] == b[i]:
                ret += 1
        return la - 1 - ret
    dp = [[0]*lb for i in range(la)]
    for i in range(1, la):
        for j in range(i, lb-la+i+1):# 这一步很重要
            if a[i] == b[j]:
                dp[i][j] = dp[i-1][j-1] + 1
            else:
                dp[i][j] = dp[i-1][j-1]
    ret = [max(dp[i]) for i in range(la)]
    ret = max(ret)
    return la - 1 - ret
 
if __name__ == '__main__':
    while 1:
        try:
            a = raw_input()
            b = raw_input()
        except:
            break
        print f(a, b)

发表于 2017-07-18 08:12:22 回复(0)
//我的思路是:将a和取长度为(a的长度)的b的连续子串作比较,看每次比较的相应位置的不同字符的个数,求最小值
let a=readline()
let b=readline()
let diff=function(a,b){
let l=a.length
let num=0
for(let i=0;i<l;i++){
if(a[i]!==b[i])num++;
}
return num;
}
let al=a.length;
let min=al;
for(let i=0,bl=b.length;i<=bl-al;i++){
let r=diff(a,b.slice(i))
if(min>r)min=r
}
print(min)

编辑于 2017-07-26 10:06:32 回复(2)
如果A是“ab”,,B是“abc”,,那A在后面添加一个“c,,不就完美啦吗?”
发表于 2017-06-14 19:19:06 回复(0)
#include <iostream>
#include <stdio.h>
#include <string.h>
 
using namespace std;
 
intdo_it(char*a, char*b){
    intlen_a = strlen(a);
    intlen_b = strlen(b);
    intc[64][64]={0};
    intmax=0;
    intm,n;
    for(inti=0;i<=len_b-len_a;i++){
        intcnt = 0;
        for(m=0,n=i;m<len_a;m++,n++){
            if(a[m]==b[n])
                cnt++;
        }
        if(cnt> max)
            max = cnt;
    }  
    returnlen_a - max;
}
 
intmain(){
    chara[64],b[64];
    cin >> a >> b;
    cout << do_it(a,b) << endl;
}

发表于 2017-06-14 16:28:00 回复(0)
#include<bits/stdc++.h>

using namespace std;

int a_match_b(string a,string b) {
int N = b.size();
int maxn = 0;
//int * match = new int[N];
//memset(match,0,sizeof(int)*N);
for (int i = 0; i < b.size()-a.size()+1;) {
int count = 0;
for (int j = 0; j < a.size(); j++,i++) {
if (a[j] == b[i])
count++;
}
//match[i] = count;
i = i - a.size()+1;
maxn = max(maxn,count);
}
return maxn;
}
int main(){
    string a, b;
    cin>>a>>b;
int min=a.size()-a_match_b(a,b);
cout << min << endl;
    return 0;
}
发表于 2017-06-09 20:59:06 回复(0)

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