给定一个字符C和一个字符串S ,任务是从字符串S 中删除第一个和最后一个出现的字符C 。
例子:
Input: S = “GeekforGeeks”, C = ‘e’
Output: GeksforGeks
Explanation:
GeekforGeeks -> GekforGeks
Input: S = “helloWorld”, C = ‘l’
Output: heloWord
方法:
思路是从两端遍历给定的字符串,找到遇到的字符C的第一次出现,并去掉对应的出现。最后,打印结果字符串。
下面是上述方法的实现:
C++
// C++ Program to implement
// the above approach
#include
using namespace std;
// Function to remove first and last
// occurrence of a given character
// from the given string
string removeOcc(string& s, char ch)
{
// Traverse the given string from
// the beginning
for (int i = 0; s[i]; i++) {
// If ch is found
if (s[i] == ch) {
s.erase(s.begin() + i);
break;
}
}
// Traverse the given string from
// the end
for (int i = s.length() - 1;
i > -1; i--) {
// If ch is found
if (s[i] == ch) {
s.erase(s.begin() + i);
break;
}
}
return s;
}
// Driver Code
int main()
{
string s = "hello world";
char ch = 'l';
cout << removeOcc(s, ch);
return 0;
}
Java
// Java Program to implement
// the above approach
class GFG{
// Function to remove first and last
// occurrence of a given character
// from the given String
static String removeOcc(String s, char ch)
{
// Traverse the given String from
// the beginning
for (int i = 0; i < s.length(); i++)
{
// If ch is found
if (s.charAt(i) == ch)
{
s = s.substring(0, i) +
s.substring(i + 1);
break;
}
}
// Traverse the given String from
// the end
for (int i = s.length() - 1; i > -1; i--)
{
// If ch is found
if (s.charAt(i) == ch)
{
s = s.substring(0, i) +
s.substring(i + 1);
break;
}
}
return s;
}
// Driver Code
public static void main(String[] args)
{
String s = "hello world";
char ch = 'l';
System.out.print(removeOcc(s, ch));
}
}
// This code is contributed by sapnasingh4991
Python3
# Python3 program to implement
# the above approach
# Function to remove first and last
# occurrence of a given character
# from the given string
def removeOcc(s, ch):
# Traverse the given string from
# the beginning
for i in range(len(s)):
# If ch is found
if (s[i] == ch):
s = s[0 : i] + s[i + 1:]
break
# Traverse the given string from
# the end
for i in range(len(s) - 1, -1, -1):
# If ch is found
if (s[i] == ch):
s = s[0 : i] + s[i + 1:]
break
return s
# Driver Code
s = "hello world"
ch = 'l'
print(removeOcc(s, ch))
# This code is contributed by sanjoy_62
C#
// C# Program to implement
// the above approach
using System;
class GFG{
// Function to remove first and last
// occurrence of a given character
// from the given String
static String removeOcc(String s, char ch)
{
// Traverse the given String from
// the beginning
for (int i = 0; i < s.Length; i++)
{
// If ch is found
if (s[i] == ch)
{
s = s.Substring(0, i) +
s.Substring(i + 1);
break;
}
}
// Traverse the given String from
// the end
for (int i = s.Length - 1; i > -1; i--)
{
// If ch is found
if (s[i] == ch)
{
s = s.Substring(0, i) +
s.Substring(i + 1);
break;
}
}
return s;
}
// Driver Code
public static void Main(String[] args)
{
String s = "hello world";
char ch = 'l';
Console.Write(removeOcc(s, ch));
}
}
// This code is contributed by sapnasingh4991
Javascript
Python3
# Python3 program to implement
# the above approach
# Function to remove first and last
# occurrence of a given character
# from the given string
def removeOcc(str, ch):
# Convert string to list
s = list(str)
# Traverse the string from starting position
for i in range(len(s)):
# if we find ch then remove it and break the loop
if(s[i] == ch):
s.pop(i)
break
# Traverse the string from the end
for i in range(len(s)-1, -1, -1):
# if we find ch then remove it and break the loop
if(s[i] == ch):
s.pop(i)
break
# Join the list
return ''.join(s)
# Driver Code
s = "hello world"
ch = 'l'
print(removeOcc(s, ch))
# This code is contributed by vikkycirus
输出:
helo word
时间复杂度: O(N)
辅助空间: O(1)
方法#2:使用索引方法
- 将字符串转换为列表。
- 遍历列表,如果我们找到给定的字符,则使用pop()删除该索引并中断循环
- 从末尾遍历列表,如果我们找到给定的字符,则使用 pop() 删除该索引并中断循环。
- 加入列表并打印它。
下面是实现:
蟒蛇3
# Python3 program to implement
# the above approach
# Function to remove first and last
# occurrence of a given character
# from the given string
def removeOcc(str, ch):
# Convert string to list
s = list(str)
# Traverse the string from starting position
for i in range(len(s)):
# if we find ch then remove it and break the loop
if(s[i] == ch):
s.pop(i)
break
# Traverse the string from the end
for i in range(len(s)-1, -1, -1):
# if we find ch then remove it and break the loop
if(s[i] == ch):
s.pop(i)
break
# Join the list
return ''.join(s)
# Driver Code
s = "hello world"
ch = 'l'
print(removeOcc(s, ch))
# This code is contributed by vikkycirus
时间复杂度: O(N)
辅助空间: O(N)
如果您希望与专家一起参加现场课程,请参阅DSA 现场工作专业课程和学生竞争性编程现场课程。