给定3个字符串S ， A和B。的任务是，以取代S的每个子串等于A与B和S的每个子串等于B，其中A。匹配A或B的两个或更多子字符串可能重叠。为避免这种情况造成混淆，您应该找到与A或B匹配的最左边的子字符串，将其替换，然后继续其余的字符串。
例如，当将A =“ aa”与S =“ aaa”匹配时， A [0，1]将优先于A [1，2] 。
请注意， A和B将具有相同的长度，并且A！= B。
例子：
Input: S = “aab”, A = “aa”, B = “bb”
Output: bbb
We match the first two characters with A and replacing it with B we get bbb.
Then we continue the algorithm starting at index 3 and we don’t find any more matches.
Input: S = “aabbaabb”, A = “aa”, B = “bb”
Output: bbaabbaa
We replace all the occurrences of “aa” with “bb” and “bb” with “aa”, so the resultant string is “bbaabbaa”.
方法：从长度为len(A)的S中遍历所有可能的子字符串。如果有任何子字符串与A或B匹配，则根据需要更新该字符串，并在最后打印更新的字符串。
下面是上述方法的实现：
C++
// C++ implementation of the approach
#include
using namespace std;
// Function to return the resultant string
string updateString(string S,
string A, string B)
{
int l = A.length();
// Iterate through all positions i
for (int i = 0; i + l <= S.length(); i++)
{
// Current sub-string of
// length = len(A) = len(B)
string curr = S.substr(i, i + l);
// If current sub-string gets
// equal to A or B
if (curr == A)
{
// Update S after replacing A
string new_string = "";
new_string += S.substr(0, i) + B +
S.substr(i + l, S.length());
S = new_string;
i += l - 1;
}
else
{
// Update S after replacing B
string new_string = "";
new_string += S.substr(0, i) + A +
S.substr(i + l, S.length());
S = new_string;
i += l - 1;
}
}
// Return the updated string
return S;
}
// Driver code
int main()
{
string S = "aab";
string A = "aa";
string B = "bb";
cout << (updateString(S, A, B)) << endl;
}
// This code is contributed by
// Surendra_Gangwar Java
// Java implementation of the approach
class GFG {
// Function to return the resultant string
static String updateString(String S, String A, String B)
{
int l = A.length();
// Iterate through all positions i
for (int i = 0; i + l <= S.length(); i++) {
// Current sub-string of length = len(A) = len(B)
String curr = S.substring(i, i + l);
// If current sub-string gets equal to A or B
if (curr.equals(A)) {
// Update S after replacing A
String new_string
= S.substring(0, i)
+ B + S.substring(i + l, S.length());
S = new_string;
i += l - 1;
}
else {
// Update S after replacing B
String new_string
= S.substring(0, i)
+ A + S.substring(i + l, S.length());
S = new_string;
i += l - 1;
}
}
// Return the updated string
return S;
}
// Driver code
public static void main(String[] args)
{
String S = "aab";
String A = "aa";
String B = "bb";
System.out.println(updateString(S, A, B));
}
}Python3
# Python3 implementation of the approach
# Function to return the resultant string
def updateString(S, A, B):
l = len(A)
# Iterate through all positions i
i = 0
while i + l <= len(S):
# Current sub-string of
# length = len(A) = len(B)
curr = S[i:i+l]
# If current sub-string gets
# equal to A or B
if curr == A:
# Update S after replacing A
new_string = S[0:i] + B + S[i + l:len(S)]
S = new_string
i += l - 1
else:
# Update S after replacing B
new_string = S[0:i] + A + S[i + l:len(S)]
S = new_string
i += l - 1
i += 1
# Return the updated string
return S
# Driver code
if __name__ == "__main__":
S = "aab"
A = "aa"
B = "bb"
print(updateString(S, A, B))
# This code is contributed by Rituraj JainC#
// C# implementation of the approach
using System;
class GFG
{
// Function to return the resultant string
static string updateString(string S, string A, string B)
{
int l = A.Length;
// Iterate through all positions i
for (int i = 0; i + l <= S.Length; i++)
{
// Current sub-string of length = len(A) = len(B)
string curr = S.Substring(i, l);
// If current sub-string gets equal to A or B
if (curr.Equals(A))
{
// Update S after replacing A
string new_string = S.Substring(0, i) +
B + S.Substring(i + l);
S = new_string;
i += l - 1;
}
else
{
// Update S after replacing B
string new_string = S.Substring(0, i) +
A + S.Substring(i + l);
S = new_string;
i += l - 1;
}
}
// Return the updated string
return S;
}
// Driver code
public static void Main()
{
string S = "aab";
string A = "aa";
string B = "bb";
Console.WriteLine(updateString(S, A, B));
}
}
// This code is contributed by RyugaPHP
输出：
bbb