未排序数组中最接近 K 的值
给定一个由 N 个整数和一个整数 K 组成的数组 arr[] ，任务是找到最接近 K 的数组元素。如果存在多个最接近的值，则打印最小的值。
示例:
输入: arr[]={4，2，8，11，7}，K = 6 输出: 7 解释: 4 和 6 的绝对差为| 4–6 | = 2 2 和 6 的绝对差为| 2–6 | = 4 8 和 6 的绝对差为| 8–6 | = 2 11 和 6 的绝对差为| 11–6 | = 5【T0 因此，K(=6)的最接近值是 7
输入: arr[]={100，200，400}，K = 300 T3】输出: 200
方法:思路是遍历给定数组并打印出与给定整数 K 的绝对差最小的数组元素。按照以下步骤解决问题:
- 初始化一个变量，比如说 res ，将数组元素存储为最接近 K 的值。
- 遍历数组，比较ABS(K–RES)的绝对值和ABS(K–arr[I])的绝对值。
- 如果ABS(K–RES)的值超过ABS(K–arr[I])，则更新 res 为 arr[i] 。
- 最后，打印 res 。
下面是上述方法的实现:
C++
// C++ program to implement
// the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to get
// the closest value
int clostVal(int arr[],
int N,
int K)
{
// Stores the closest
// value to K
int res = arr[0];
// Traverse the array
for (int i = 1; i < N;
i++) {
// If absolute difference
// of K and res exceeds
// absolute difference of K
// and current element
if (abs(K - res) > abs(K - arr[i])) {
res = arr[i];
}
}
// Return the closest
// array element
return res;
}
// Driver Code
int main()
{
int arr[] = { 100, 200, 400 };
int K = 300;
int N = sizeof(arr) / sizeof(arr[0]);
cout << clostVal(arr, N, K);
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java program to implement
// the above approach
import java.io.*;
class GFG{
// Function to get
// the closest value
static int clostVal(int arr[], int N,
int K)
{
// Stores the closest
// value to K
int res = arr[0];
// Traverse the array
for(int i = 1; i < N; i++)
{
// If absolute difference
// of K and res exceeds
// absolute difference of K
// and current element
if (Math.abs(K - res) >
Math.abs(K - arr[i]))
{
res = arr[i];
}
}
// Return the closest
// array element
return res;
}
// Driver Code
public static void main (String[] args)
{
int arr[] = { 100, 200, 400 };
int K = 300;
int N = arr.length;
System.out.print(clostVal(arr, N, K));
}
}
// This code is contributed by code_hunt
Python 3
# Python3 program to implement
# the above approach
# Function to get
# the closest value
def clostVal(arr, N, K):
# Stores the closest
# value to K
res = arr[0]
# Traverse the array
for i in range(1, N, 1):
# If absolute difference
# of K and res exceeds
# absolute difference of K
# and current element
if (abs(K - res) >
abs(K - arr[i])):
res = arr[i]
# Return the closest
# array element
return res
# Driver Code
arr = [ 100, 200, 400 ]
K = 300
N = len(arr)
print(clostVal(arr, N, K))
# This code is contributed by susmitakundugoaldanga
C
// C# program to implement
// the above approach
using System;
class GFG{
// Function to get
// the closest value
static int clostVal(int[] arr, int N,
int K)
{
// Stores the closest
// value to K
int res = arr[0];
// Traverse the array
for(int i = 1; i < N; i++)
{
// If absolute difference
// of K and res exceeds
// absolute difference of K
// and current element
if (Math.Abs(K - res) >
Math.Abs(K - arr[i]))
{
res = arr[i];
}
}
// Return the closest
// array element
return res;
}
// Driver Code
public static void Main ()
{
int[] arr = { 100, 200, 400 };
int K = 300;
int N = arr.Length;
Console.WriteLine(clostVal(arr, N, K));
}
}
// This code is contributed by sanjoy_62
java 描述语言
<script>
// Javascript program to implement
// the above approach
// Function to get
// the closest value
function clostVal(arr, N, K)
{
// Stores the closest
// value to K
let res = arr[0];
// Traverse the array
for(let i = 1; i < N; i++)
{
// If absolute difference
// of K and res exceeds
// absolute difference of K
// and current element
if (Math.abs(K - res) >
Math.abs(K - arr[i]))
{
res = arr[i];
}
}
// Return the closest
// array element
return res;
}
// Driver Code
let arr = [ 100, 200, 400 ];
let K = 300;
let N = arr.length;
document.write(clostVal(arr, N, K));
</script>
Output:
200
时间复杂度:O(N) T5辅助空间:** O(1)
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