给定一个数组，任务是从给定数组的元素按升序创建一个新的排序数组。
例子：
Input : arr[] = {2, 5, 4, 9, 8}
Output : 2 4 5 8 9
Input : arr[] = {10, 45, 98, 35, 45}
Output : 10 35 45 45 98使用二分搜索可以有效地解决上述问题。我们创建一个新数组并在它为空时插入第一个元素。现在对于每个新元素，我们使用二分搜索找到新数组中元素的正确位置，然后将该元素插入新数组中的相应索引处。
下面是上述方法的实现：
C++
// C++ program to create a sorted array
// using Binary Search
#include
using namespace std;
// Function to create a new sorted array
// using Binary Search
void createSorted(int a[], int n)
{
// Auxiliary Array
vector b;
for (int j = 0; j < n; j++) {
// if b is empty any element can be at
// first place
if (b.empty())
b.push_back(a[j]);
else {
// Perform Binary Search to find the correct
// position of current element in the
// new array
int start = 0, end = b.size() - 1;
// let the element should be at first index
int pos = 0;
while (start <= end) {
int mid = start + (end - start) / 2;
// if a[j] is already present in the new array
if (b[mid] == a[j]) {
// add a[j] at mid+1. you can add it at mid
b.emplace(b.begin() + max(0, mid + 1), a[j]);
break;
}
// if a[j] is lesser than b[mid] go right side
else if (b[mid] > a[j])
// means pos should be between start and mid-1
pos = end = mid - 1;
else
// else pos should be between mid+1 and end
pos = start = mid + 1;
// if a[j] is the largest push it at last
if (start > end) {
pos = start;
b.emplace(b.begin() + max(0, pos), a[j]);
// here max(0, pos) is used because sometimes
// pos can be negative as smallest duplicates
// can be present in the array
break;
}
}
}
}
// Print the new generated sorted array
for (int i = 0; i < n; i++)
cout << b[i] << " ";
}
// Driver Code
int main()
{
int a[] = { 2, 5, 4, 9, 8 };
int n = sizeof(a) / sizeof(a[0]);
createSorted(a, n);
return 0;
} Java
// Java program to create a sorted array
// using Binary Search
import java.util.*;
class GFG
{
// Function to create a new sorted array
// using Binary Search
static void createSorted(int a[], int n)
{
// Auxiliary Array
Vector b = new Vector<>();
for (int j = 0; j < n; j++)
{
// if b is empty any element can be at
// first place
if (b.isEmpty())
b.add(a[j]);
else
{
// Perform Binary Search to find the correct
// position of current element in the
// new array
int start = 0, end = b.size() - 1;
// let the element should be at first index
int pos = 0;
while (start <= end)
{
int mid = start + (end - start) / 2;
// if a[j] is already present in the new array
if (b.get(mid) == a[j])
{
// add a[j] at mid+1. you can add it at mid
b.add((Math.max(0, mid + 1)), a[j]);
break;
}
// if a[j] is lesser than b[mid] go right side
else if (b.get(mid) > a[j])
// means pos should be between start and mid-1
pos = end = mid - 1;
else
// else pos should be between mid+1 and end
pos = start = mid + 1;
// if a[j] is the largest push it at last
if (start > end)
{
pos = start;
b.add(Math.max(0, pos), a[j]);
// here max(0, pos) is used because sometimes
// pos can be negative as smallest duplicates
// can be present in the array
break;
}
}
}
}
// Print the new generated sorted array
for (int i = 0; i < n; i++)
System.out.print(b.get(i) + " ");
}
// Driver Code
public static void main(String args[])
{
int a[] = { 2, 5, 4, 9, 8 };
int n = a.length;
createSorted(a, n);
}
}
/* This code is contributed by PrinciRaj1992 */ Python3
# Python program to create a sorted array
# using Binary Search
# Function to create a new sorted array
# using Binary Search
def createSorted(a: list, n: int):
# Auxiliary Array
b = []
for j in range(n):
# if b is empty any element can be at
# first place
if len(b) == 0:
b.append(a[j])
else:
# Perform Binary Search to find the correct
# position of current element in the
# new array
start = 0
end = len(b) - 1
# let the element should be at first index
pos = 0
while start <= end:
mid = start + (end - start) // 2
# if a[j] is already present in the new array
if b[mid] == a[j]:
# add a[j] at mid+1. you can add it at mid
b.insert(max(0, mid + 1), a[j])
break
# if a[j] is lesser than b[mid] go right side
elif b[mid] > a[j]:
# means pos should be between start and mid-1
pos = end = mid - 1
else:
# else pos should be between mid+1 and end
pos = start = mid + 1
# if a[j] is the largest push it at last
if start > end:
pos = start
b.insert(max(0, pos), a[j])
# here max(0, pos) is used because sometimes
# pos can be negative as smallest duplicates
# can be present in the array
break
# Print the new generated sorted array
for i in range(n):
print(b[i], end=" ")
# Driver Code
if __name__ == "__main__":
a = [2, 5, 4, 9, 8]
n = len(a)
createSorted(a, n)
# This code is contributed by
# sanjeev2552C#
// C# program to create a sorted array
// using Binary Search
using System;
using System.Collections.Generic;
class GFG
{
// Function to create a new sorted array
// using Binary Search
static void createSorted(int []a, int n)
{
// Auxiliary Array
List b = new List();
for (int j = 0; j < n; j++)
{
// if b is empty any element can be at
// first place
if (b.Count == 0)
b.Add(a[j]);
else
{
// Perform Binary Search to find the correct
// position of current element in the
// new array
int start = 0, end = b.Count - 1;
// let the element should be at first index
int pos = 0;
while (start <= end)
{
int mid = start + (end - start) / 2;
// if a[j] is already present in the new array
if (b[mid] == a[j])
{
// add a[j] at mid+1. you can add it at mid
b.Insert((Math.Max(0, mid + 1)), a[j]);
break;
}
// if a[j] is lesser than b[mid] go right side
else if (b[mid] > a[j])
// means pos should be between start and mid-1
pos = end = mid - 1;
else
// else pos should be between mid+1 and end
pos = start = mid + 1;
// if a[j] is the largest push it at last
if (start > end)
{
pos = start;
b.Insert(Math.Max(0, pos), a[j]);
// here Max(0, pos) is used because sometimes
// pos can be negative as smallest duplicates
// can be present in the array
break;
}
}
}
}
// Print the new generated sorted array
for (int i = 0; i < n; i++)
Console.Write(b[i] + " ");
}
// Driver Code
public static void Main(String []args)
{
int []a = { 2, 5, 4, 9, 8 };
int n = a.Length;
createSorted(a, n);
}
}
// This code is contributed by 29AjayKumar Javascript
输出：
2 4 5 8 9时间复杂度：O(N*N)。尽管正在使用二进制搜索，但列表插入调用平均运行时间为 O(N)
辅助空间：O(N)
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