给定两个长度为N 的字符串X和Y ，任务是检查是否可以通过将X 的任何子字符串反转一次来使这两个字符串相等。如果可能，则打印“是” 。否则，打印“否” 。
例子：
Input: X = “adcbef”, Y = “abcdef”
Output: Yes
Explanation: Strings can be made equal by reversing the substring “dcb” of string X.
Input: X = “126543”, Y = “123456”
Output: Yes
Explanation: Strings can be made equal by reversing the substring “6543” of string X.
朴素的方法：解决问题的最简单的方法是反转字符串X 的每个可能的子串，并且对于每个反转，检查两个字符串是否相等。如果在任何逆转后发现为真，则打印“是” 。否则，打印“否”。
时间复杂度： O(N 2 )
辅助空间： O(1)
高效方法：对上述方法进行优化，请按照以下步骤解决问题：
- 初始化一个变量，比如L为-1 ，以存储两个字符串具有不相等字符的左侧第一个索引。
- 使用变量i在[0, N – 1]范围内遍历字符串X ，如果对于任何索引，如果发现两个字符串中的字符不相等，则设置L = i并跳出循环。
- 初始化一个变量，比如R为-1 ，以存储从右边开始的第一个索引，在两个字符串具有不相等的字符。
- 使用变量i在[N – 1, 0]范围内遍历字符串X ，如果对于任何索引，如果发现两个字符串中的字符不相等，则设置R = i并跳出循环。
- 在索引[L, R] 上反转字符串X的字符。
- 完成上述步骤后，检查两个字符串是否相等。如果发现相等，则打印“Yes” 。否则，打印“否” 。
下面是上述方法的实现：
C++
// C++ program for the above approach
#include
using namespace std;
// Function to check if the strings
// can be made equal or not by
// reversing a substring of X
bool checkString(string X, string Y)
{
// Store the first index from
// the left which contains unequal
// characters in both the strings
int L = -1;
// Store the first element from
// the right which contains unequal
// characters in both the strings
int R = -1;
// Checks for the first index from
// left in which characters in both
// the strings are unequal
for (int i = 0; i < X.length(); ++i) {
if (X[i] != Y[i]) {
// Store the current index
L = i;
// Break out of the loop
break;
}
}
// Checks for the first index from
// right in which characters in both
// the strings are unequal
for (int i = X.length() - 1; i > 0; --i) {
if (X[i] != Y[i]) {
// Store the current index
R = i;
// Break out of the loop
break;
}
}
// Reverse the substring X[L, R]
reverse(X.begin() + L,
X.begin() + R + 1);
// If X and Y are equal
if (X == Y) {
cout << "Yes";
}
// Otherwise
else {
cout << "No";
}
}
// Driver Code
int main()
{
string X = "adcbef", Y = "abcdef";
// Function Call
checkString(X, Y);
return 0;
} Java
// Java program for the above approach
import java.util.*;
class GFG{
// Function to check if the Strings
// can be made equal or not by
// reversing a subString of X
static void checkString(String X, String Y)
{
// Store the first index from
// the left which contains unequal
// characters in both the Strings
int L = -1;
// Store the first element from
// the right which contains unequal
// characters in both the Strings
int R = -1;
// Checks for the first index from
// left in which characters in both
// the Strings are unequal
for (int i = 0; i < X.length(); ++i) {
if (X.charAt(i) != Y.charAt(i)) {
// Store the current index
L = i;
// Break out of the loop
break;
}
}
// Checks for the first index from
// right in which characters in both
// the Strings are unequal
for (int i = X.length() - 1; i > 0; --i) {
if (X.charAt(i) != Y.charAt(i)) {
// Store the current index
R = i;
// Break out of the loop
break;
}
}
// Reverse the subString X[L, R]
X = X.substring(0, L) +
reverse(X.substring(L, R + 1)) +
X.substring(R + 1);
// If X and Y are equal
if (X.equals(Y)) {
System.out.print("Yes");
}
// Otherwise
else {
System.out.print("No");
}
}
static String reverse(String input) {
char[] a = input.toCharArray();
int l, r = a.length - 1;
for (l = 0; l < r; l++, r--) {
char temp = a[l];
a[l] = a[r];
a[r] = temp;
}
return String.valueOf(a);
}
// Driver Code
public static void main(String[] args)
{
String X = "adcbef", Y = "abcdef";
// Function Call
checkString(X, Y);
}
}
// This code is contributed by 29AjayKumarPython3
# Python3 program for the above approach
# Function to check if the strings
# can be made equal or not by
# reversing a substring of X
def checkString(X, Y):
# Store the first index from
# the left which contains unequal
# characters in both the strings
L = -1
# Store the first element from
# the right which contains unequal
# characters in both the strings
R = -1
# Checks for the first index from
# left in which characters in both
# the strings are unequal
for i in range(len(X)):
if (X[i] != Y[i]):
# Store the current index
L = i
# Break out of the loop
break
# Checks for the first index from
# right in which characters in both
# the strings are unequal
for i in range(len(X) - 1, 0, -1):
if (X[i] != Y[i]):
# Store the current index
R = i
# Break out of the loop
break
X = list(X)
X = X[:L] + X[R : L - 1 : -1 ] + X[R + 1:]
# If X and Y are equal
if (X == list(Y)):
print("Yes")
# Otherwise
else:
print("No")
# Driver Code
if __name__ == "__main__" :
X = "adcbef"
Y = "abcdef"
# Function Call
checkString(X, Y)
# This code is contributed by AnkThonC#
// C# program for the above approach
using System;
class GFG{
// Function to check if the Strings
// can be made equal or not by
// reversing a subString of X
static void checkString(String X, String Y)
{
// Store the first index from
// the left which contains unequal
// characters in both the Strings
int L = -1;
// Store the first element from
// the right which contains unequal
// characters in both the Strings
int R = -1;
// Checks for the first index from
// left in which characters in both
// the Strings are unequal
for(int i = 0; i < X.Length; ++i)
{
if (X[i] != Y[i])
{
// Store the current index
L = i;
// Break out of the loop
break;
}
}
// Checks for the first index from
// right in which characters in both
// the Strings are unequal
for(int i = X.Length - 1; i > 0; --i)
{
if (X[i] != Y[i])
{
// Store the current index
R = i;
// Break out of the loop
break;
}
}
// Reverse the subString X[L, R]
X = X.Substring(0, L) +
reverse(X.Substring(L, R + 1 - L)) +
X.Substring(R + 1);
// If X and Y are equal
if (X.Equals(Y))
{
Console.Write("Yes");
}
// Otherwise
else
{
Console.Write("No");
}
}
static String reverse(String input)
{
char[] a = input.ToCharArray();
int l, r = a.Length - 1;
for(l = 0; l < r; l++, r--)
{
char temp = a[l];
a[l] = a[r];
a[r] = temp;
}
return String.Join("",a);
}
// Driver Code
public static void Main(String[] args)
{
String X = "adcbef", Y = "abcdef";
// Function Call
checkString(X, Y);
}
}
// This code is contributed by Amit Katiyar输出：
Yes时间复杂度： O(N)
辅助空间： O(1)
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