在排序数组中搜索 K 的插入位置

给定一个由N 个不同整数和一个整数K组成的排序数组arr[] ，任务是找到 K 的索引，如果它存在于数组arr[] 中。否则，找到必须插入K以保持数组排序的索引。

例子：

Input: arr[] = {1, 3, 5, 6}, K = 5
Output: 2
Explanation: Since 5 is found at index 2 as arr[2] = 5, the output is 2.

Input: arr[] = {1, 3, 5, 6}, K = 2
Output: 1
Explanation: Since 2 is not present in the array but can be inserted at index 1 to make the array sorted.

朴素的方法：按照以下步骤解决问题：

• 迭代数组arr[] 的每个元素并搜索整数K 。
• 如果发现任何数组元素等于K ，则打印K 的索引。
• 否则，如果发现任何数组元素大于K ，则将该索引打印为K的插入位置。如果没有发现超过K 的元素，则必须在最后一个数组元素之后插入K。

以下是上述方法的实现：

C++

// C++ program for the above approach
 
#include 
using namespace std;
 
// Function to find insert position of K
int find_index(int arr[], int n, int K)
{
    // Traverse the array
    for (int i = 0; i < n; i++)
 
        // If K is found
        if (arr[i] == K)
            return i;
 
        // If current array element
        // exceeds K
        else if (arr[i] > K)
            return i;
 
    // If all elements are smaller
    // than K
    return n;
}
 
// Driver Code
int main()
{
    int arr[] = { 1, 3, 5, 6 };
    int n = sizeof(arr) / sizeof(arr[0]);
    int K = 2;
    cout << find_index(arr, n, K) << endl;
    return 0;
}

Java

// Java program for the above approach
import java.io.*;
 
class GFG{
 
// Function to find insert position of K
static int find_index(int[] arr, int n, int K)
{
     
    // Traverse the array
    for(int i = 0; i < n; i++)
     
        // If K is found
        if (arr[i] == K)
            return i;
 
        // If current array element
        // exceeds K
        else if (arr[i] > K)
            return i;
 
    // If all elements are smaller
    // than K
    return n;
}
 
// Driver Code
public static void main(String[] args)
{
    int[] arr = { 1, 3, 5, 6 };
    int n = arr.length;
    int K = 2;
     
    System.out.println(find_index(arr, n, K));
}
}
 
// This code is contributed by akhilsaini

Python3

# Python program for the above approach
 
# Function to find insert position of K
def find_index(arr, n, K):
     
    # Traverse the array
    for i in range(n):
         
        # If K is found
        if arr[i] == K:
            return i
             
        # If arr[i] exceeds K
        elif arr[i] > K:
            return i
             
    # If all array elements are smaller
    return n
 
# Driver Code
arr = [1, 3, 5, 6]
n = len(arr)
K = 2
print(find_index(arr, n, K))

C#

// C# program for the above approach
using System;
 
class GFG{
 
// Function to find insert position of K
static int find_index(int[] arr, int n, int K)
{
     
    // Traverse the array
    for(int i = 0; i < n; i++)
     
        // If K is found
        if (arr[i] == K)
            return i;
 
        // If current array element
        // exceeds K
        else if (arr[i] > K)
            return i;
 
    // If all elements are smaller
    // than K
    return n;
}
 
// Driver Code
public static void Main()
{
    int[] arr = { 1, 3, 5, 6 };
    int n = arr.Length;
    int K = 2;
     
    Console.WriteLine(find_index(arr, n, K));
}
}
 
// This code is contributed by akhilsaini

Javascript

 
// Javascript program for the above approach
 
// Function to find insert position of K
function find_index(arr, n, K)
{
     
    // Traverse the array
    for(let i = 0; i < n; i++)
      
        // If K is found
        if (arr[i] == K)
            return i;
  
        // If current array element
        // exceeds K
        else if (arr[i] > K)
            return i;
  
    // If all elements are smaller
    // than K
    return n;
}
 
// Driver code
let arr = [ 1, 3, 5, 6 ];
let n = arr.length;
let K = 2;
  
document.write(find_index(arr, n, K));
 
// This code is contributed by splevel62  
 

输出：

1

时间复杂度： O(N)
辅助空间： O(1)

高效的方法：为了优化上述方法，想法是使用二分搜索。请按照以下步骤解决问题：

• 将start和end设置为0和N – 1 ，其中start和end变量分别表示搜索空间的下限和上限。
• 计算mid = (start + end) / 2 。
• 如果发现arr[mid]等于K ，则打印mid作为所需答案。
• 如果arr[mid]超过K ，则设置low = mid + 1 。否则，设置high = mid – 1 。

以下是上述方法的实现：

C++

// C++ program for the above approach
 
#include 
using namespace std;
 
// Function to find insert position of K
int find_index(int arr[], int n, int K)
{
    // Lower and upper bounds
    int start = 0;
    int end = n - 1;
 
    // Traverse the search space
    while (start <= end) {
        int mid = (start + end) / 2;
 
        // If K is found
        if (arr[mid] == K)
            return mid;
 
        else if (arr[mid] < K)
            start = mid + 1;
 
        else
            end = mid - 1;
    }
 
    // Return insert position
    return end + 1;
}
 
// Driver Code
int main()
{
    int arr[] = { 1, 3, 5, 6 };
    int n = sizeof(arr) / sizeof(arr[0]);
    int K = 2;
    cout << find_index(arr, n, K) << endl;
    return 0;
}

Java

// Java program for the above approach
import java.io.*;
 
class GFG{
 
// Function to find insert position of K
static int find_index(int[] arr, int n, int K)
{
     
    // Lower and upper bounds
    int start = 0;
    int end = n - 1;
 
    // Traverse the search space
    while (start <= end)
    {
        int mid = (start + end) / 2;
 
        // If K is found
        if (arr[mid] == K)
            return mid;
 
        else if (arr[mid] < K)
            start = mid + 1;
 
        else
            end = mid - 1;
    }
 
    // Return insert position
    return end + 1;
}
 
// Driver Code
public static void main(String[] args)
{
    int[] arr = { 1, 3, 5, 6 };
    int n = arr.length;
    int K = 2;
     
    System.out.println(find_index(arr, n, K));
}
}
 
// This code is contributed by akhilsaini

蟒蛇3

# Python program to implement
# the above approach
 
# Function to find insert position of K
def find_index(arr, n, B):
 
    # Lower and upper bounds
    start = 0
    end = n - 1
 
    # Traverse the search space
    while start<= end:
 
        mid =(start + end)//2
 
        if arr[mid] == K:
            return mid
 
        elif arr[mid] < K:
            start = mid + 1
        else:
            end = mid-1
 
    # Return the insert position
    return end + 1
 
# Driver Code
arr = [1, 3, 5, 6]
n = len(arr)
K = 2
print(find_index(arr, n, K))

C＃

// C# program for the above approach
using System;
 
class GFG{
 
// Function to find insert position of K
static int find_index(int[] arr, int n, int K)
{
     
    // Lower and upper bounds
    int start = 0;
    int end = n - 1;
 
    // Traverse the search space
    while (start <= end)
    {
        int mid = (start + end) / 2;
 
        // If K is found
        if (arr[mid] == K)
            return mid;
 
        else if (arr[mid] < K)
            start = mid + 1;
 
        else
            end = mid - 1;
    }
 
    // Return insert position
    return end + 1;
}
 
// Driver Code
public static void Main()
{
    int[] arr = { 1, 3, 5, 6 };
    int n = arr.Length;
    int K = 2;
     
    Console.WriteLine(find_index(arr, n, K));
}
}
 
// This code is contributed by akhilsaini

Javascript

// JavaScript program for the above approach
 
// Function to find insert position of K
function find_index(arr, n, K)
{
    // Lower and upper bounds
    let start = 0;
    let end = n - 1;
 
    // Traverse the search space
    while (start <= end) {
        let mid = Math.floor((start + end) / 2);
 
        // If K is found
        if (arr[mid] == K)
            return mid;
 
        else if (arr[mid] < K)
            start = mid + 1;
 
        else
            end = mid - 1;
    }
 
    // Return insert position
    return end + 1;
}
 
// Driver Code
    let arr = [ 1, 3, 5, 6 ];
    let n = arr.length;
    let K = 2;
    document.write(find_index(arr, n, K) + "<br>");
 
 
 
 
// This code is contributed by Surbhi Tyagi.

输出：

1

时间复杂度： O(log N)
辅助空间： O(1)

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