Question

来自interviewstreet.com的

* 问题（字符串排列） *

给定两个字符串，编写一个方法来判断一个是否是一个排列的另一个。您的解决方案应考虑区分大小写和空白是重要的。
一组对象的排列是这些对象的排列特定的顺序。例如，有六种排列字符串“abc”，即“abc”，“acb”，“bac”，“bca”，“cab”和 “CBA”。
输出：如果两个字符串是彼此的排列，则返回1。如果两个字符串不是彼此的排列，则返回0。
abc acb cab cba bac bca

import java.io。*;

import java.util。*;

公共类解决方案{

私人集置换;

public static void main(String args[] ) throws Exception {
 //   Scanner sc = new Scanner(System.in);
    //String string1 = sc.nextLine();
    //String string2 = sc.nextLine();

String string1 = "str";
    String string2 = "str";

    Solution solution = new Solution();
    int output = solution.permutation(string1, string2);
    System.out.println(output);
}

public  void stringPermuation(String s1, String s2) {

if (s2.length() > 0) {

for (int i = 0; i < s2.length(); i++) {
    System.out.println(s1 + s2.charAt(i)+","+ s2.substring(0, i)+" +"+ s2.substring(i + 1));

    stringPermuation(s1 + s2.charAt(i),
        s2.substring(0, i) + s2.substring(i + 1));
    }}
else{
    permutations.add(s1);
    System.out.println(s1);
}
}

public Set stringPermuation(String s) {
permutations = new HashSet<String>();
stringPermuation("", s);
return permutations;
}



private  int permutation(String string1, String string2) {
int result = 0;
 Set<String> setString1 = stringPermuation(string1);
 Set<String> setString2 = stringPermuation(string2);
    // create an iterator

     System.out.println("There are total of " + setString1.size() + " permutations in String1:");
     System.out.println("There are total of " + setString2.size() + " permutations in String2:");

     if(setString1.size() == setString2.size())
        result=IterateSet(setString1,setString2);

    //Return 1 if string1 is a permutation of string2
    //Return 0 if string1 is not a permutation of string2
return result;
}

public  int IterateSet(Set setString1,Set setString2){
    int  i=  0;
    Iterator<String> it = setString1.iterator();
    while (it.hasNext()) {
        if(setString2.contains(it.next()) && i == 0)
         i=1;
    }
return i;}}

Answer 1

...对两个字符串的字符进行排序（按字典顺序排列），如果两个排序的字符串相等，则原始是彼此的排列。

（String Permutation）来自interviewstreet.com

1 个答案: