不使用库函数检查两个字符串是否相同

原文:https://www . geeksforgeeks . org/check-如果两个字符串相同或不相同，则不使用库函数/

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。

示例:

输入:S1 =“geesforgeks”，S2 =“geesforgeks” T3】输出:T5】真 T7】说明:T9】S1 和 S2 是同一个字符串

输入: S1 =【极客 ForGeeks】，S2 =【极客 forgeeks】 T3】输出:T5】假

方法:按照以下步骤解决问题:

  • 创建一个函数comparistrings()，该函数将两个字符串 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，则返回假。
    • 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】，则返回假。
    • 遍历结束时返回真。

下面是上述方法的实现:

C

#include <stdbool.h>
#include <stdio.h>

// 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;
}

Python 3

# 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

java 描述语言

<script>

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
</script>

Output

True

时间复杂度:O(N) T5辅助空间:** O(1)

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