将元素数组连接成单个元素
给定一个由 N 个整数组成的数组， arr[] ，任务是打印通过连接数组元素获得的单个整数值。
示例:
输入: arr[] = {1，23，345 } T3】输出: 12345
输入: arr[] = {123，45，6，78 } T3】输出: 12345678
方法:给定的问题可以基于以下观察来解决:
- Consider x and y as two integer values to be connected. And the length of the integer y is also considered as l .
- Then two integers x and y can be connected together as follows:
- X×10 l +Y
按照以下步骤解决问题:
- 初始化一个变量，说 ans 为 0，存储结果值。
- 使用变量 i、遍历数组 arr[] ，然后在每次迭代中将 ans 乘以 10 乘以的幂整数 arr[i] 中的位数，并将 ans 增加 arr[i]。
- 最后，完成上述步骤后，打印在和中获得的答案。
下面是上述方法的实现:
C++
// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to find the integer value
// obtained by joining array elements
// together
int ConcatenateArr(int arr[], int N)
{
// Stores the resulting integer value
int ans = arr[0];
// Traverse the array arr[]
for (int i = 1; i < N; i++) {
// Stores the count of digits of
// arr[i]
int l = floor(log10(arr[i]) + 1);
// Update ans
ans = ans * pow(10, l);
// Increment ans by arr[i]
ans += arr[i];
}
// Return the ans
return ans;
}
// Driver Code
int main()
{
// Input
int arr[] = { 1, 23, 456 };
int N = sizeof(arr) / sizeof(arr[0]);
// Function call
cout << ConcatenateArr(arr, N);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java program for the above approach
import java.util.*;
class GFG{
// Function to find the integer value
// obtained by joining array elements
// together
static int ConcatenateArr(int[] arr, int N)
{
// Stores the resulting integer value
int ans = arr[0];
// Traverse the array arr[]
for(int i = 1; i < N; i++)
{
// Stores the count of digits of
// arr[i]
int l = (int)Math.floor(Math.log10(arr[i]) + 1);
// Update ans
ans = ans * (int)Math.pow(10, l);
// Increment ans by arr[i]
ans += arr[i];
}
// Return the ans
return ans;
}
// Driver Code
public static void main(String args[])
{
// Input
int arr[] = { 1, 23, 456 };
int N = arr.length;
// Function call
System.out.println(ConcatenateArr(arr, N));
}
}
// This code is contributed by avijitmondal1998
Python 3
# Python3 program for the above approach
import math
# Function to find the integer value
# obtained by joining array elements
# together
def ConcatenateArr(arr, N):
# Stores the resulting integer value
ans = arr[0]
# Traverse the array arr[]
for i in range(1, N):
# Stores the count of digits of
# arr[i]
l = math.floor(math.log10(arr[i]) + 1)
# Update ans
ans = ans * math.pow(10, l)
# Increment ans by arr[i]
ans += arr[i]
# Return the ans
return int( ans)
# Driver Code
if __name__ == "__main__":
# Input
arr = [1, 23, 456]
N = len(arr)
# Function call
print(ConcatenateArr(arr, N))
# This code is contributed by ukasp.
C
// C# program for the above approach
using System;
class GFG{
// Function to find the integer value
// obtained by joining array elements
// together
static int ConcatenateArr(int[] arr, int N)
{
// Stores the resulting integer value
int ans = arr[0];
// Traverse the array arr[]
for(int i = 1; i < N; i++)
{
// Stores the count of digits of
// arr[i]
int l = (int)Math.Floor(Math.Log10(arr[i]) + 1);
// Update ans
ans = ans * (int)Math.Pow(10, l);
// Increment ans by arr[i]
ans += arr[i];
}
// Return the ans
return ans;
}
// Driver Code
public static void Main()
{
// Input
int[] arr = { 1, 23, 456 };
int N = arr.Length;
// Function call
Console.Write(ConcatenateArr(arr, N));
}
}
// This code is contributed by sanjoy_62.
java 描述语言
<script>
// JavaScript program for the above approach
// Function to find the integer value
// obtained by joining array elements
// together
function ConcatenateArr(arr, N)
{
// Stores the resulting integer value
let ans = arr[0];
// Traverse the array arr[]
for (let i = 1; i < N; i++) {
// Stores the count of digits of
// arr[i]
let l = Math.floor(Math.log10(arr[i]) + 1);
// Update ans
ans = ans * Math.pow(10, l);
// Increment ans by arr[i]
ans += arr[i];
}
// Return the ans
return ans;
}
// Driver Code
// Input
let arr = [1, 23, 456];
let N = arr.length;
// Function call
document.write(ConcatenateArr(arr, N));
// This code is contributed by Potta Lokesh
</script>
Output
123456
时间复杂度: O(Nlog(M))，其中 M 是数组的最大元素。* 辅助空间: O(1)
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