Java中的字符串度量方法

Question

我正在解决用Java处理字符串的问题。请帮我解决这个问题。

问题的情况:约翰启动了他的新公司来识别他所见过的云，他称之为长度为N的字符串A。但突然间，他发现山姆也推出了自己的云识别初创公司，并将其命名为string B of length N。

更正式地说，让我们来看看字符串A (约翰的创业公司的名字)和string B (山姆的创业公司的名字)。两个字符串都是相同长度的N。对于字符串B的每个位置1≤i≤N，您需要使用string A计算该位置的匹配类型。

If a
Bi=Ai, then in position i the match type must be equal to P (from the word plagiarism).

If Bi≠Ai, but there is another position 1≤j≤N such that Bi=Aj, then in position i
  match type must be equal to S (from the word suspicious).

注：

 Letters within one line can be repeated.
 Each letter of string A can be used in at most one plagiarism or suspicious match.
 Preference is always given to the plagiarism type.
 In the case of a suspicious match, the leftmost position in row A is always preferred.

。在其他位置，匹配类型必须等于i(从单词innocent)。

输入格式

The first line contains the string

A(1≤∣∣A∣∣≤10^6) is the startup name chosen by John.

The second line contains the string B(|B|=|A|) — the name of Sam's startup.

It is guaranteed that strings A and B
  contain only uppercase latin letters.

输出格式

Output a single line
C(|C|=|B|), where Ci is the match type of the character Bi(1≤i≤|B|):
for type plagiarism Ci= P.

for type suspicious Ci=S.

for type innocentCi=I.

示例1:

Input      Output

CLOUD     PSIIP
CUPID

示例2:

Input      Output

ALICE     SPII
ELIBO

示例3:

Input       Output

ABCBCYA    IPSSPIP
ZBBACAA

备注：

Explanation for the first test


B1=A1 and B5=A5 , so for positions 1 and 5 the answer is P.

B2≠A2 , but B2=A4, so for position 2 the answer is S.

Letters P and I do not occur in string A, so for positions 3 and 4 the answer is I.


Explanation for the second test:


B2=A2 and B3=A3, so for positions 2 and 3 the answer is P.

B1≠A1 , but B1=A5, so for position 1 the answer is S.

Letters B and O do not occur in string A, so for positions 4 and 5 the answer is I.


Explanation for the third test:


B2=A2 , B5=A5 and B7=A7 so for positions 2, 5 and 7 the answer is P.

B3≠A3 but B3=A2=A4. A2 is already enabled according to B2=A2,

therefore, the correspondence B3=A4 is chosen - for position 3 the answer is S.

B4≠A4 and B6≠A6, but B4=B6=A1=A7.

A7 is already enabled according to B7=A7;

4<6, therefore, for position 4, the correspondence B4=A1 (answer S) is selected;

there are no matches left for position 6 (answer I).

The letter Z does not occur in string A, so for position 1 the answer is I.

A7 is already enabled according to B7=A7;

4<6, therefore, for position 4, the correspondence B4=A1 (answer S) is selected;

there are no matches left for position 6 (answer I).

The letter Z does not occur in string A, so for position 1 the answer is I.

我的解决方案：

public class Solution {
  public boolean backspaceCompare(String s, String t) {
    return formBackSpaceString(s).equals(formBackSpaceString(t));
  }
  
  private String formBackSpaceString(String s) {
    Stack<Character> stack = new Stack<>();
    for (char c : s.toCharArray()) {
      if (c == '#') {
        if (!stack.isEmpty()) {
          stack.pop();
        }
      } else {
        stack.push(c);
      }
    }
    StringBuilder sb = new StringBuilder();
    while (!stack.isEmpty()) {
      sb.append(stack.pop());
    }
    return sb.toString();
  }
public static void main (String[] args){
 }
}

我深陷于这项任务的逻辑之中。我将非常感谢帮助，至少在伪码级别。

Answer 1

您提供的代码似乎与您描述的问题无关。

这个问题要求从原始字符串中应用一个简单的字符规则(Ai，Bi)来构造一个新的字符串C。只有三条规则：

• ，如果Ai == Bi，Ci =“P”
• ，如果A包含char Bi，Ci =“S”
• ，Ci = "I"

您可以使用流API以一种简单的方式做到这一点：

String problem(String A, String B) {
    // construct a set of all chars in A
    Set<Integer> aChars = A.chars().boxed().collect(Collectors.toSet());

    // apply rules for chars Ai, Bi
    return IntStream.range(0, A.length())
            .mapToObj(i -> A.charAt(i) == B.charAt(i) ? "P" :
                    aChars.contains((int) B.charAt(i)) ? "S" : "I")
            .collect(Collectors.joining());
}

或者，更详细地说，没有它：

String problem(String A, String B) {
    Set<Character> aChars = new HashSet<>();
    for (int i = 0; i < A.length(); i++) {
        aChars.add(A.charAt(i));
    }

    StringBuilder builder = new StringBuilder();
    for (int i = 0; i < A.length(); i++) {
        if (A.charAt(i) == B.charAt(i)) {
            builder.append("P");
        } else if (aChars.contains(B.charAt(i))) {
            builder.append("S");
        } else {
            builder.append("I");
        }
    }
    return builder.toString();
}