不使用库函数检查两个字符串是否相同

给定两个字符串S1和S2 ，任务是在不使用字符串库函数的情况下检查它们是否相同。

例子：

Input: S1 = ”GeeksForGeeks”, S2 = ”GeeksForGeeks”
Output: 
True
Explanation:
S1 and S2 are the same strings

Input: S1 = ”GeeksForGeeks”, S2 = ”GeeksforGeeks”
Output: 
False

解决方法：按照以下步骤解决问题：

• 创建一个函数compareStrings() ，它将两个字符串S1和S2作为输入参数并执行以下操作：
  • 如果S1和S2的长度不同，则返回false 。
  • 将S1的长度存储在一个变量中，比如N 。
  • 使用变量i从0遍历到N-1并执行以下操作：
    • 如果S1[i]不等于S2[i] ，则返回false 。
  • 在遍历结束时返回true 。

下面是上述方法的实现：

C

#include 
#include 
 
// Function to calculate length of string
int len(char* S)
{
    // Variable for traversal
    int i = 0;
   
    // Traverse till null is reached
    while (S[i])
        i++;
   
    return i;
}
 
// Function to check whether
// two strings are same or not
bool compareStrings(char S1[], char S2[])
{
    // If lengths of the two
      // strings are different
    if (len(S1) != len(S2))
        return false;
   
    // Variable for traversal
    int i = 0;
   
    // Traverse till null is reached
    while (S1[i]) {
       
        if (S1[i] != S2[i])
            return false;
       
          // Increment i
        i++;
    }
    return true;
}
 
// Driver Code
int main()
{
    // Input
    char S1[] = "GeeksForGeeks";
    char S2[] = "GeeksForGeeks";
 
    // Function Call
    bool ans = compareStrings(S1, S2);
   
    printf("%s", ans ? "True" : "False");
  
    return 0;
}

Python3

# Function to check whether
# two strings are same or not
def compareStrings(S1, S2):
     
    # If lengths of the two
    # strings are different
    if (len(S1) != len(S2)):
        return False
 
    # Variable for traversal
    i = 0
 
    # Traverse till null is reached
    while (i < len(S1)):
        if (S1[i] != S2[i]):
            return False
 
        # Increment i
        i += 1
         
    return True
 
# Driver Code
if __name__ == '__main__':
 
    # Input
    S1 = "GeeksForGeeks"
    S2 = "GeeksForGeeks"
 
    # Function Call
    ans = compareStrings(S1, S2)
 
    print("True" if ans else "False")
 
# This code is contributed by mohit kumar 29

C#

using System;
using System.Collections.Generic;
 
class GFG{
  
// Function to calculate length of string
static int len(string S)
{
    // Variable for traversal
    int i = 0;
   
    // Traverse till null is reached
    while (i < S.Length)
        i++;
   
    return i;
}
 
// Function to check whether
// two strings are same or not
static bool compareStrings(string S1, string S2)
{
     
    // If lengths of the two
    // strings are different
    if (len(S1) != len(S2))
        return false;
   
    // Variable for traversal
    int i = 0;
   
    // Traverse till null is reached
    while (i < S1.Length)
    {
        if (S1[i] != S2[i])
            return false;
             
        // Increment i
        i++;
    }
    return true;
}
 
// Driver Code
public static void Main()
{
     
    // Input
    string S1 = "GeeksForGeeks";
    string S2 = "GeeksForGeeks";
 
    // Function Call
    bool ans = compareStrings(S1, S2);
   
    Console.Write(ans ? "True" : "False");
}
}
     
// This code is contributed by SURENDRA_GANGWAR

Javascript

 
function len(S)
{
    // Variable for traversal
    let i = 0;
    
    // Traverse till null is reached
    while (i < S.length)
        i++;
    
    return i;
}
 
function compareStrings(S1,S2)
{
    // If lengths of the two
    // strings are different
    if (len(S1) != len(S2))
        return false;
    
    // Variable for traversal
    let i = 0;
    
    // Traverse till null is reached
    while (i < S1.length)
    {
        if (S1[i] != S2[i])
            return false;
              
        // Increment i
        i++;
    }
    return true;
}
 
// Input
let S1 = "GeeksForGeeks";
let S2 = "GeeksForGeeks";
 
// Function Call
let ans = compareStrings(S1, S2);
 
document.write(ans ? "True" : "False");
 
 
// This code is contributed by patel2127

输出

True

时间复杂度： O(N)
辅助空间： O(1)