编程找出两个字符串中最大公共子字符串,如"abccade",

#include<iostream>
#include<stdio.h>
#include<cmath>
#include<algorithm>
using namespace std;
 
int GetCommon(char s1[], char s2[])
{
    int len1=strlen(s1);
    int len2=strlen(s2);
    int r,maxlen=0;
    
    for(int i=0;i<len1;i++)
    {
        for(int j=0;j<len2;j++)
        {
            if(s1[i]==s2[j])
            {
                int as=i,bs=j,count=1;
                while(as+1<len1&&bs+1<len2 &&s1[++as]==s2[++bs])
                    count++;
                if(count>maxlen)
                {
                    maxlen = count;
                    r=i;
                }
            }
        }
    }
    for(int i=r;i<r+maxlen;i++)
        printf("%c",s1[i]);
    printf("\n");
    return maxlen;
}
int main()
{
    char a[]={"abccade"};
    char b[]={"dgcadde"};
    printf("%s和%s的最长公共子串是：\n",a,b);
    printf("长度为：%d\n",GetCommon(a,b));
    
    return 0;
}

class Solution {
    public String comString(String strA, String strB) {
        char[] a = strA.toCharArray();
        char[] b = strB.toCharArray();
        StringBuilder common =  new StringBuilder();
        String preCom = "";

        int i = 0, j = 0;
        while (i < a.length && j < b.length) {
            // 在b里找一个字符跟a[i]相同，直至找到或到头儿
            while (j < b.length && a[i] != b[j]) {
                j++;
            }
            // 回溯存档点
            int index = i;
            while (i < a.length && j < b.length && a[i] == b[j]) {
                common.append(a[i]);
                i++;
                j++;
            }
            // 如果这一次找到的字符串比上一次长，更新
            if (preCom.length() < common.length()) {
                preCom = common.toString();
            }
            common = new StringBuilder();
            // 回溯
            i = index + 1;
            j = 0;
        }
        return preCom;
    }
}    

a = input('第一个字符串：')
b = input('第二个字符串：')
m = [[0 for i in range(len(b)+1)] for j in range(len(a)+1)]
mmax = 0
p = 0
for i in range(len(a)):         for j in range(len(b)):             if a[i]==b[j]:                 m[i+1][j+1]=m[i][j]+1                 if m[i+1][j+1]>mmax:                     mmax=m[i+1][j+1]                     p=i+1
print (a[p-mmax:mmax])

#include "iostream"
using namespace std;

struct record//此结构体用于记录历史上最大的子串信息
{     int i;     int j;     int size;
public:     record(int a, int b, int c)     {         i = a;         j = b;         size = c;     }     void remember(int a, int b, int c)     {         i = a;         j = b;         size = c;     }
};
int _tmain(int argc, _TCHAR* argv[])
{
        //输入部分     char* str1 = new char[100];     char* str2 = new char[100];     fstream fin("10.txt");     fin >> str1;     fin >> str2;     int strlen1 = strlen(str1);     int strlen2 = strlen(str2);     record rec(0, 0, 0);     for (int i = 0; i < strlen1 && (strlen1 - i) >= rec.size;i++)     {//当横向遍历剩下的字符数目小于历史记录中最大子串的长度时候 则没有必要遍历之后的所有字符         for (int j = 0; j < strlen2 && (strlen2 - j) >= rec.size; j++)         {
//当纵向遍历剩下的字符数目小于历史记录中最大子串的长度时候 则没有必要遍历之后的所有字符
            int ti = i;

            int tj = j;

            int count = 0;

            while (str1[ti]!='\0'&&str2[tj]!='\0'&&str1[ti] == str2[tj])

            {

                count++;

                ti++;

                tj++;

            }

            if (count > rec.size)

            {

                rec.remember(i, j, count);

            }

        }

    }

    

    int k = rec.i;

    for (int i = 0; i < rec.size; i++, k++)

    {

        cout << str1[k];

        

    }

    return 0;

}

import java.util.Scanner;

public class Main {
    public static void main(String[] args) {
        Scanner scanner=new Scanner(System.in);
        String s1=scanner.nextLine();
        String s2=scanner.nextLine();
        //找出两个字符串中谁的长度长
        String min=(s1.length()<=s2.length())?s1:s2;
        String max=(min.equals(s1))?s2:s1;
        //如果长的字符串包含短的字符串，直接返回
        if(max.contains(min)){
            System.out.println(min);
            return;
        }
        //字符串从末尾开始慢慢缩减长度，如果包含就直接输出并返回即可
        //外层循环控制start的起始位置
        for(int i=0;i<min.length();i++){
            //每次都是从末端缩减长度，这样的话，如果包含就可以直接输出，即为包含的最长的字符串
            for(int start=i,end=min.length();end>start+1;end--){
                String temp=min.substring(start,end);
                if(max.contains(temp)){
                    System.out.println(temp);
                    return;
                }
            }
        }
        
    }
}

private static String getSameStringInfo(String strFirst,String strSecond){
        int strLengthFirst = strFirst.length();
        String finallyStr = null;
        int lengthFlag = 0;
        for (int i = 0; i < strLengthFirst; i++) {
            for (int j = i+1; j <= strLengthFirst; j++) {
                String newStr = strFirst.substring(i,j);
                if (strSecond.contains(newStr) && newStr.length() > lengthFlag){
                    lengthFlag = newStr.length();
                    finallyStr = newStr;
                }
            }
        }
        return finallyStr;
    }

	#include<iostream>
#include<algorithm>
#include<vector>
#include<string>



	using namespace std;



	string longestCommonSubstring(string &A, string &B) 
{
 int Alen = A.size();
 int Blen = B.size();
 if (Alen == 0 || Blen == 0)
 {
  return 0;
 }
 int max = 0;
 int flag = 0;
 for (int i = 0; i < Alen; i++)
 {
  for (int j = 0; j < Blen; j++)
  {
   int m = i, n = j;
   int len = 0;
   while (m < Alen && n < Blen)
   {
    if (A[m] != B[n])
    {
     break;
    }
    m++;
    n++;
    len++;
   }
   if (len > max)
   {
    flag = n;
    max = len;
   }
  }
 }
 return A.substr(flag - max, flag - 1);
}



	int main()
{
 string s1, s2;
 cin >> s1 >> s2;
 string res = longestCommonSubstring(s1, s2);
 cout << res << endl;
 return 0;
}

public static String getSubString(String s1,String s2) { //找出长度较小的字符串  String min = s1.length()<s2.length()?s1:s2;  String max = min.equals(s1)?s2:s1;  //字符串的截取和比较  for(int i=0; i<min.length(); i++){ //控制前后下标的移动  int j=0, k=min.length()-i;  while(k <= min.length()){
            String sub = min.substring(j,k);  if(max.contains(sub)){ return sub;  }
            j++;  k++;  }
    } return null; }