包含 C2 的最长子串，以 C1 开始，以 C3 结束
给定一个字符串S 。查找以字符C1开头、以字符C3结尾并且其间至少有一个字符C2的最长子字符串。如果不存在这样的子字符串，则打印“-1” 。
注意：大写和小写字母的处理方式不同，因此“a”和“A”不等价。
例子：
Input: S = “aasdfdsafsdasf”, C1 = ‘d’, C2 = ‘s’, C3 = ‘d’
Output: 8 dfdsafsd
Explanation: “dfdsafsd” starts with ‘d’ and ends with ‘d’ and two ‘s’ in between.
Input: S = “GeeksForGeeks”, C1 = ‘G’, C2 = ‘e’, C3 = ‘k’
Output: 12 GeeksForGeek
Explanation: “GeeksForGeek” starts with ‘G’ and ends with ‘k’ and four ‘e’ in between.
Input: S = “Prophecy”, C1 = ‘p’, C2 = ‘p’, C3 = ‘t’
Output: -1
Explanation: No substring exists with starting character as ‘p’ and ending as ‘p’.
方法：可以使用以下想法解决问题：
Find the first occurrence of C1 and last occurrence of C3. If there is any C2 in between them, then this is the longest possible substring. Otherwise, no such substring is possible.
请按照以下步骤解决问题：
- 在S中找到C1的第一次出现（比如i ）。
- 从i开始迭代到字符串的结尾：
- 将字符附加到最长的子字符串。
- 如果任何字符与C2的C2递增计数相同。
- 如果找到C3并且C2的计数不为零，则更新最长的子字符串。
- 如果计数为 0，则返回 -1。
- 否则，返回最后一次出现C3之前形成的子字符串。
下面是上述方法的实现。
C++14
// C++ code to implement the approach
#include
using namespace std;
// Function to find the longest substring
string Solve(string& S, char& C1,
char& C2, char& C3)
{
int maxLen = 0, countB = 0,
i, j, n = S.length();
// First occurrence of C1
for (i = 0; i < n; i++) {
if (S[i] == C1) {
j = i;
break;
}
}
// Finding the longest substring
while (j++ < n) {
if (S[j] == C2)
countB++;
if (countB > 0 && S[j] == C3)
maxLen = max(maxLen, j - i + 1);
}
if (maxLen == 0)
return "-1";
return S.substr(i, maxLen);
}
// Driver code
int main()
{
string S = "GeeksForGeeks";
char C1 = 'G';
char C2 = 'e';
char C3 = 'k';
string ans = Solve(S, C1, C2, C3);
if (ans.compare("-1") == 0)
cout << "-1";
else
cout << ans.length() << " "
<< ans;
return 0;
} Java
// Java code to implement the approach
public class GFG
{
// Function to find the longest substring
static String Solve(String S, char C1, char C2, char C3)
{
int maxLen = 0;
int countB = 0;
int i, j = 0;
int n = S.length();
// First occurrence of C1
for (i = 0; i < n; i++) {
if (S.charAt(i) == C1) {
j = i;
break;
}
}
// Finding the longest substring
while (++j < n) {
if (S.charAt(j) == C2)
countB++;
if (countB > 0 && S.charAt(j) == C3)
maxLen = Math.max(maxLen, j - i + 1);
}
if (maxLen == 0)
return "-1";
return S.substring(i, maxLen);
}
// Driver code
public static void main(String[] args)
{
String S = "GeeksForGeeks";
char C1 = 'G';
char C2 = 'e';
char C3 = 'k';
String ans = Solve(S, C1, C2, C3);
if (ans == "-1")
System.out.println("-1");
else
System.out.println(ans.length() + " " + ans);
}
}
//This code is contributed by phasing17Python3
# Python 3 code to implement the approach
# Function to find the longest substrin
def Solve(S, C1,
C2, C3):
maxLen = 0
countB = 0
n = len(S)
# First occurrence of C1
for i in range(n):
if (S[i] == C1):
j = i
break
# Finding the longest substring
while (j < n):
if (S[j] == C2):
countB += 1
if (countB > 0 and S[j] == C3):
maxLen = max(maxLen, j - i + 1)
j += 1
if (maxLen == 0):
return "-1"
return S[i: maxLen]
# Driver code
if __name__ == "__main__":
S = "GeeksForGeeks"
C1 = 'G'
C2 = 'e'
C3 = 'k'
ans = Solve(S, C1, C2, C3)
if (ans == "-1"):
print("-1")
else:
print(len(ans), ans)
# This code is contributed by ukasp.C#
// C# code to implement the approach
using System;
class GeeksForGeeks {
// Function to find the longest substring
static string Solve(string S, char C1, char C2, char C3)
{
int maxLen = 0, countB = 0, i = 0, j = 0,
n = S.Length;
// First occurrence of C1
for (i = 0; i < n; i++) {
if (S[i] == C1) {
j = i;
break;
}
}
// Finding the longest substring
while (j < n) {
if (S[j] == C2)
countB++;
if (countB > 0 && S[j] == C3)
maxLen = Math.Max(maxLen, j - i + 1);
j++;
}
if (maxLen == 0)
return "-1";
return S.Substring(i, maxLen);
}
// Driver code
public static void Main()
{
string S = "GeeksForGeeks";
char C1 = 'G';
char C2 = 'e';
char C3 = 'k';
string ans = Solve(S, C1, C2, C3);
if (ans == "-1")
Console.Write("-1");
else
Console.Write(ans.Length + " " + ans);
}
}
// This code is contributed by Samim Hossain Mondal.Javascript
12 GeeksForGeek时间复杂度： O(N)，其中 N 是字符串的长度
辅助空间： O(1)