给定两个字符串S1和S2 ，任务是通过交换任意一对字符替换字符串S1中的任何字符来检查字符串S1是否等于字符串S2 。如果可以使字符串S1等于S2 ，则打印“是” 。否则，打印“否” 。
例子：
Input: S1 = “abc”, S2 = “bca”
Output: Yes
Explanation:
Operation 1: “abc” -> “acb”
Operation 2: “acb” -> “bca”
Input: S1 = “a”, S2 = “aa”
Output: No
Explanation: It is impossible to make the two strings same.
方法：解决此问题的想法是找到字符串字符的频率，并检查两个字符串是否使用了相同的字符。请按照以下步骤解决问题：
- 初始化两个数组a []和b []以存储字符串S1和S2的频率。
- 遍历字符串S1和S2并将字符串的频率存储在数组a []和b []中。
- 对数组a []和b []进行排序。
- 遍历两个数组，如果任何元素在任何索引处都不相等，则打印“否”，因为两个字符串中至少有一个字符不同。
- 完成上述步骤后，如果不存在不相等的频率，则可以打印“是”，因为可以通过给定的操作将字符串S1转换为字符串S2 。
以下是此方法的实现：
C++
// C++ program for the above approach
#include
using namespace std;
// Function to find if given strings
// are same or not
bool sameStrings(string str1,
string str2)
{
int N = str1.length();
int M = str2.length();
// Base Condition
if (N != M) {
return false;
}
// Stores frequency of characters
// of the string str1 and str2
int a[256] = { 0 }, b[256] = { 0 };
// Traverse strings str1 & str2 and
// store frequencies in a[] and b[]
for (int i = 0; i < N; i++) {
a[str1[i] - 'a']++;
b[str2[i] - 'a']++;
}
// Check if both strings have
// same characters or not
int i = 0;
while (i < 256) {
if ((a[i] == 0 && b[i] == 0)
|| (a[i] != 0 && b[i] != 0)) {
i++;
}
// If a character is present
// in one string and is not in
// another string, return false
else {
return false;
}
}
// Sort the array a[] and b[]
sort(a, a + 256);
sort(b, b + 256);
// Check arrays a and b contain
// the same frequency or not
for (int i = 0; i < 256; i++) {
// If the frequencies are not
// the same after sorting
if (a[i] != b[i])
return false;
}
// At this point, str1 can
// be converted to str2
return true;
}
// Driver Code
int main()
{
string S1 = "cabbba", S2 = "abbccc";
if (sameStrings(S1, S2))
cout << "YES" << endl;
else
cout << " NO" << endl;
return 0;
} Java
// Java program for the above approach
import java.util.*;
class GFG
{
// Function to find if given Strings
// are same or not
static boolean sameStrings(String str1,
String str2)
{
int N = str1.length();
int M = str2.length();
// Base Condition
if (N != M)
{
return false;
}
// Stores frequency of characters
// of the String str1 and str2
int []a = new int[256];
int []b = new int[256];
// Traverse Strings str1 & str2 and
// store frequencies in a[] and b[]
for (int i = 0; i < N; i++)
{
a[str1.charAt(i) - 'a']++;
b[str2.charAt(i) - 'a']++;
}
// Check if both Strings have
// same characters or not
int i = 0;
while (i < 256)
{
if ((a[i] == 0 && b[i] == 0)
|| (a[i] != 0 && b[i] != 0))
{
i++;
}
// If a character is present
// in one String and is not in
// another String, return false
else
{
return false;
}
}
// Sort the array a[] and b[]
Arrays.sort(a);
Arrays.sort(b);
// Check arrays a and b contain
// the same frequency or not
for (i = 0; i < 256; i++)
{
// If the frequencies are not
// the same after sorting
if (a[i] != b[i])
return false;
}
// At this point, str1 can
// be converted to str2
return true;
}
// Driver Code
public static void main(String[] args)
{
String S1 = "cabbba", S2 = "abbccc";
if (sameStrings(S1, S2))
System.out.print("YES" +"\n");
else
System.out.print(" NO" +"\n");
}
}
// This code is contributed by 29AjayKumarPython3
# Python3 program for the above approach
# Function to find if given strings
# are same or not
def sameStrings(str1, str2):
N = len(str1)
M = len(str2)
# Base Condition
if (N != M):
return False
# Stores frequency of characters
# of the str1 and str2
a, b = [0]*256, [0]*256
# Traverse strings str1 & str2 and
# store frequencies in a[] and b[]
for i in range(N):
a[ord(str1[i]) - ord('a')] += 1
b[ord(str2[i]) - ord('a')] += 1
# Check if both strings have
# same characters or not
i = 0
while (i < 256):
if ((a[i] == 0 and b[i] == 0) or (a[i] != 0 and b[i] != 0)):
i += 1
# If a character is present
# in one and is not in
# another string, return false
else:
return False
# Sort the array a[] and b[]
a = sorted(a)
b = sorted(b)
# Check arrays a and b contain
# the same frequency or not
for i in range(256):
# If the frequencies are not
# the same after sorting
if (a[i] != b[i]):
return False
# At this point, str1 can
# be converted to str2
return True
# Driver Code
if __name__ == '__main__':
S1, S2 = "cabbba", "abbccc"
if (sameStrings(S1, S2)):
print("YES")
else:
print("NO")
# This code is contributed by mohit kumar 29.C#
// C# program for the above approach
using System;
class GFG
{
// Function to find if given Strings
// are same or not
static bool sameStrings(string str1,
string str2)
{
int N = str1.Length;
int M = str2.Length;
// Base Condition
if (N != M)
{
return false;
}
// Stores frequency of characters
// of the String str1 and str2
int []a = new int[256];
int []b = new int[256];
// Traverse Strings str1 & str2 and
// store frequencies in a[] and b[]
for (int j = 0; j < N; j++)
{
a[str1[j] - 'a']++;
b[str2[j] - 'a']++;
}
// Check if both Strings have
// same characters or not
int i = 0 ;
while (i < 256)
{
if ((a[i] == 0 && b[i] == 0)
|| (a[i] != 0 && b[i] != 0))
{
i++;
}
// If a character is present
// in one String and is not in
// another String, return false
else
{
return false;
}
}
// Sort the array a[] and b[]
Array.Sort(a);
Array.Sort(b);
// Check arrays a and b contain
// the same frequency or not
for (int j = 0; j < 256; j++)
{
// If the frequencies are not
// the same after sorting
if (a[j] != b[j])
return false;
}
// At this point, str1 can
// be converted to str2
return true;
}
// Driver Code
static public void Main()
{
string S1 = "cabbba", S2 = "abbccc";
if (sameStrings(S1, S2))
Console.Write("YES" +"\n");
else
Console.Write(" NO" +"\n");
}
}
// This code is contributed by code_hunt.输出：
YES时间复杂度： O(N)
辅助空间： O(256)