跪求算法！！！

求算法。给定2个4位的各个位数上的数字都不重复的数字，如果他们位置和数字都相等，为A，如果数字相等位置不等，为B。例如：
给定的第1个数：5673
第2个数： 0153
这样匹配的结果为：1A1B
变换第2个数各个数字的位置：5013
这样匹配的结果为2A0B
不知道大家明白我的意思没有，跪求该算法！！！


[解决办法]
楼主可以求出每个数的个十百千，然后用一个数的个十百千循环去和另一个数的个十百千比较，然后计算相同或者不同的个数，不知道楼主明白了吗？楼主可以自己整理算法
[解决办法]

Java code

import java.io.IOException;import java.util.Scanner;public class Main {    public static void main(String[] args) throws IOException {        String num1, num2;        // user String to save **** int        Scanner in = new Scanner(System.in);        num1 = in.nextLine();        num2 = in.nextLine();        int A = 0, B = 0;        for (int i = 0; i < 4; i++) {            for (int j = 0; j < 4; j++) {                if (num1.charAt(i) == num2.charAt(j))                    B++;            }        }        for (int i = 0; i < 4; i++) {            if (num1.charAt(i) == num2.charAt(i))                A++;        }        System.out.println(A + "A" + (B - A) + "B");    }}
[解决办法]
Java code
public static void main(String[] args) {        int a = 5673;        int b = 5013;        HashMap list = new HashMap();        while (a > 0) {            list.put(list.size(), a % 10);            a /= 10;        }        int k1 = 0, k2 = 0;        int i = 0;        while (b > 0) {            int t = b % 10;            if (list.containsValue(t)) {                if (list.get(i) == Integer.valueOf(t))                    k1++;                else                    k2++;            }            i++;            b /= 10;        }        System.out.println(k1 + "A" + k2 + "B");    }
[解决办法]
Java code
      int A = 0;      int B = 0;      String s1 = "5673";      String s2 = "0153";      if(s1.length() == s2.length()) {         for(int i = 0; i < s1.length(); i++) {            char c1 = s1.charAt(i);            for(int j = 0; j < s2.length(); j++) {               char c2 = s2.charAt(j);               if(c1 == c2) {                  if(i == j) {                     A++;                  }                  else {                     B++;                  }               }            }         }      }      System.err.println(A + "A" + B + "B");
[解决办法]
Java code
    public static void main(String[] args){        int a = 0, b = 0;        String s1 = "5673";        String s2 = "5013";        for (int j = 0, i = 0; i<s1.length();)        {            if (s1.charAt(i)==s2.charAt(j)){                if (i==j){                    a++;                }                else{                    b++;                }            }            j++;            if (j==s1.length()){                i++;                j = 0;            }        }        System.out.println(a+"A"+b+"B");    }
[解决办法]
Java code
char [] num1 = new Integer(5673).toString().toCharArray();        char [] num2 = new Integer(5013).toString().toCharArray();        int counterA = 0, counterB = 0;        for (int i = 0; i < 4; i++) {            for (int j = 0; j < 4; j++) {                if (j!=i&&num1[i] == num2[j])                    counterB++;            }            if (num1[i] == num2[i] )counterA++;                        }        System.out.println(counterA + "A" + counterB + "B"); 
 
[解决办法]
算法不会，只会 code
对于两个位数不重复的可以这样
Java code
public String t3(){        int A = 0;        int B = 0;        int num1 = 1345;        int num2 = 1534;                if(num1 == num2){            A = 4;            B = 0;            return A+"A"+B+"B";        }                String str1 = "" + num1;        String str2 = "" + num2;        char[] chars1 = str1.toCharArray();        char[] chars2 = str2.toCharArray();        for(int i =0 ; i< chars1.length;i++){            int c = chars1[i];            if(str2.indexOf(c) != -1){//有此                if(str2.charAt(i) == c){                    A ++ ;                }else{                    B ++;                }            }        }                        return A+"A"+B+"B";    }
[解决办法]
Java code
  public static void main(String[] args) throws Exception {    Scanner in = new Scanner(System.in);    for (;;) {      String num1 = in.nextLine();      String num2 = in.nextLine();      int mask1 = 0;      int mask2 = 0;      int a = 0;      for (int i = 0; i < 4; i++) {        int n1 = num1.charAt(i) - '0';        int n2 = num2.charAt(i) - '0';        mask1 |= (1 << n1);        mask2 |= (1 << n2);        if (n1 == n2) {          a++;        }      }      int b = 0;      for (int i = 0; i < 10; i++, mask1 >>= 1, mask2 >>= 1) {        int m1 = mask1 & 1;        int m2 = mask2 & 1;        if (m1 + m2 == 2) {          b++;        }      }      System.out.printf("%dA%dB%n", a, b);    }  }
[解决办法]
使用二进制计数器，等价于数组计数器
Java code
    Scanner in = new Scanner(System.in);    for (;;) {      String num1 = in.nextLine();      String num2 = in.nextLine();      int[] count1 = new int[10];      int[] count2 = new int[10];      int a = 0;      for (int i = 0; i < 4; i++) {        int n1 = num1.charAt(i) - '0';        int n2 = num2.charAt(i) - '0';        count1[n1]++;        count2[n2]++;        if (n1 == n2) {          a++;        }      }      int b = 0;      for (int i = 0; i < 10; i++) {        if (count1[i] + count2[i] == 2) {          b++;        }      }      System.out.printf("%dA%dB%n", a, b);    }