给定两个字符串，先检查哪个字符串构成回文
原文:https://www . geesforgeks . org/given-two-string-check-string-makes-回文-first/
给定两个长度相等的字符串“A”和“B”。两个玩家玩一个游戏，他们都从各自的字符串中选择一个字符(第一个从 A 中选择，第二个从 B 中选择)，并放入第三个字符串(最初是空的)。能做出第三串回文的玩家，就是赢家。如果第一个玩家先做回文，然后打印“A”，否则打印“B”。如果字符串变空，没有人能组成回文，那么打印“B”。 例:
Input : A = ab
B = ab
Output : B
First player puts 'a' (from string A)
Second player puts 'a' (from string B)
which make palindrome.
The result would be same even if A picks
'b' as first character.
Input : A = aba
B = cde
Output : A
Input : A = ab
B = cd
Output : B
None of the string will be able to
make a palindrome (of length > 1)
in any situation. So B will win.
举几个例子后，我们可以观察到‘A’(或第一个玩家)只有当它有一个角色出现不止一次，并且没有出现在‘B’中时，才能获胜。
C++
// Given two strings, check which string
// makes palindrome first.
#include<bits/stdc++.h>
using namespace std;
const int MAX_CHAR = 26;
// returns winner of two strings
char stringPalindrome(string A, string B)
{
// Count frequencies of characters in
// both given strings
int countA[MAX_CHAR] = {0};
int countB[MAX_CHAR] = {0};
int l1 = A.length(), l2 = B.length();
for(int i=0; i<l1;i++)
countA[A[i]-'a']++;
for(int i=0; i<l2;i++)
countB[B[i]-'a']++;
// Check if there is a character that
// appears more than once in A and does
// not appear in B
for (int i=0 ;i <26;i++)
if ((countA[i] >1 && countB[i] == 0))
return 'A';
return 'B';
}
// Driver Code
int main()
{
string a = "abcdea";
string b = "bcdesg";
cout << stringPalindrome(a,b);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java program to check which string
// makes palindrome first.
public class First_Palin {
static final int MAX_CHAR = 26;
// returns winner of two strings
static char stringPalindrome(String A, String B)
{
// Count frequencies of characters in
// both given strings
int[] countA = new int[MAX_CHAR];
int[] countB = new int[MAX_CHAR];
int l1 = A.length();
int l2 = B.length();
for (int i = 0; i < l1; i++)
countA[A.charAt(i) - 'a']++;
for (int i = 0; i < l2; i++)
countB[B.charAt(i) - 'a']++;
// Check if there is a character that
// appears more than once in A and does
// not appear in B
for (int i = 0; i < 26; i++)
if ((countA[i] > 1 && countB[i] == 0))
return 'A';
return 'B';
}
// Driver Code
public static void main(String args[])
{
String a = "abcdea";
String b = "bcdesg";
System.out.println(stringPalindrome(a, b));
}
}
// This code is contributed by Sumit Ghosh
Python 3
# Given two strings, check which string
# makes palindrome first.
MAX_CHAR = 26
# returns winner of two strings
def stringPalindrome(A, B):
# Count frequencies of characters
# in both given strings
countA = [0] * MAX_CHAR
countB = [0] * MAX_CHAR
l1 = len(A)
l2 = len(B)
for i in range(l1):
countA[ord(A[i]) - ord('a')] += 1
for i in range(l2):
countB[ord(B[i]) - ord('a')] += 1
# Check if there is a character that
# appears more than once in A and
# does not appear in B
for i in range(26):
if ((countA[i] > 1 and countB[i] == 0)):
return 'A'
return 'B'
# Driver Code
if __name__ == '__main__':
a = "abcdea"
b = "bcdesg"
print(stringPalindrome(a, b))
# This code is contributed by Rajput-Ji
C
// C# program to check which string
// makes palindrome first.
using System;
class First_Palin {
static int MAX_CHAR = 26;
// returns winner of two strings
static char stringPalindrome(string A, string B)
{
// Count frequencies of characters in
// both given strings
int[] countA = new int[MAX_CHAR];
int[] countB = new int[MAX_CHAR];
int l1 = A.Length;
int l2 = B.Length;
for (int i = 0; i < l1; i++)
countA[A[i] - 'a']++;
for (int i = 0; i < l2; i++)
countB[B[i] - 'a']++;
// Check if there is a character that
// appears more than once in A and does
// not appear in B
for (int i = 0; i < 26; i++)
if ((countA[i] > 1 && countB[i] == 0))
return 'A';
return 'B';
}
// Driver Code
public static void Main()
{
string a = "abcdea";
string b = "bcdesg";
Console.WriteLine(stringPalindrome(a, b));
}
}
// This code is contributed by vt_m.
服务器端编程语言（Professional Hypertext Preprocessor 的缩写）
<?php
// Given two strings, check which string
// makes palindrome first.
$MAX_CHAR = 26;
// returns winner of two strings
function stringPalindrome($A, $B)
{
global $MAX_CHAR;
// Count frequencies of characters in
// both given strings
$countA = array_fill(0, $MAX_CHAR, 0);
$countB = array_fill(0, $MAX_CHAR, 0);
$l1 = strlen($A);
$l2 = strlen($B);
for($i = 0; $i < $l1; $i++)
$countA[ord($A[$i])-ord('a')]++;
for($i = 0; $i < $l2; $i++)
$countB[ord($B[$i])-ord('a')]++;
// Check if there is a character that
// appears more than once in A and does
// not appear in B
for ($i = 0 ; $i < 26; $i++)
if (($countA[$i] > 1 && $countB[$i] == 0))
return 'A';
return 'B';
}
// Driver Code
$a = "abcdea";
$b = "bcdesg";
echo stringPalindrome($a,$b);
// This code is contributed by mits
?>
java 描述语言
<script>
// javascript program to check which string
// makes palindrome first.
var MAX_CHAR = 26;
// returns winner of two strings
function stringPalindrome(A, B)
{
// Count frequencies of characters in
// both given strings
var countA = Array.from({length: MAX_CHAR}, (_, i) => 0);
var countB = Array.from({length: MAX_CHAR}, (_, i) => 0);
var l1 = A.length;
var l2 = B.length;
for (var i = 0; i < l1; i++)
countA[A.charAt(i).charCodeAt(0) - 'a'.charCodeAt(0)]++;
for (var i = 0; i < l2; i++)
countB[B.charAt(i).charCodeAt(0) - 'a'.charCodeAt(0)]++;
// Check if there is a character that
// appears more than once in A and does
// not appear in B
for (var i = 0; i < 26; i++)
if ((countA[i] > 1 && countB[i] == 0))
return 'A';
return 'B';
}
// Driver Code
var a = "abcdea";
var b = "bcdesg";
document.write(stringPalindrome(a, b));
// This code is contributed by 29AjayKumar
</script>
输出:
A
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