生成范围【L，R】内所有长度相同的二进制数
原文:https://www . geesforgeks . org/generate-all-binary-numbers-in-range-l-r-同长/
给定两个正整数 L 和 R 。任务是把 L 到 R 的所有数字转换成二进制数。所有二进制数的长度应该相同。
示例:
输入: L = 2，R = 4 输出: 010 011 100 说明:数字的二进制表示:2 = 10，3 = 11，4 = 100。 对于长度相同的数字，在 3 和 4 的二进制表示中加一个 0。
输入: L = 2，R = 8 输出: 0010 0011 0100 0101 0110 0111 1000
方法:按照下面提到的方法解决问题。
- 要确定结果二进制数的长度，取 R+1 的对数为基数 2。
- 然后从 L 到 R 遍历，将每个数字转换成确定长度的二进制。
- 存储每个数字，并在最后打印出来。
下面是上述方法的实现
C++
// C++ code to implement the approach
#include <bits/stdc++.h>
using namespace std;
// Function to convert a number to binary
vector<int> convertToBinary(int num,
int length)
{
vector<int> bits(length, 0);
if (num == 0) {
return bits;
}
int i = length - 1;
while (num != 0) {
bits[i--] = (num % 2);
// Integer division
// gives quotient
num = num / 2;
}
return bits;
}
// Function to convert all numbers
// in range [L, R] to binary of
// same length
vector<vector<int> > getAllBinary(int l,
int r)
{
// Length of the binary numbers
int n = (int) ceil(log(r+1) / log (2));
vector<vector<int> > binary_nos;
for (int i = l; i <= r; i++) {
vector<int> bits =
convertToBinary(i, n);
binary_nos.push_back(bits);
}
return binary_nos;
}
// Driver code
int main()
{
int L = 2, R = 8;
vector<vector<int> > binary_nos =
getAllBinary(L, R);
for (int i = 0; i < binary_nos.size();
i++) {
for (int j = 0; j <
binary_nos[i].size(); j++)
cout << binary_nos[i][j];
cout << endl;
}
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java code to implement the approach
import java.util.*;
public class GFG
{
// Function to convert a number to binary
static ArrayList<Integer> convertToBinary(int num,
int length)
{
ArrayList<Integer> bits= new ArrayList<Integer>();
for(int i = 0; i < length; i++) {
bits.add(0);
}
if (num == 0) {
return bits;
}
int i = length - 1;
while (num != 0) {
bits.set(i, (num % 2));
i = i - 1;
// Integer division
// gives quotient
num = num / 2;
}
return bits;
}
// Function to convert all numbers
// in range [L, R] to binary of
// same length
static ArrayList<ArrayList<Integer> > getAllBinary(int l,
int r)
{
// Length of the binary numbers
double x = Math.log(r+1);
double y = Math.log (2);
int n = (int) Math.ceil(x / y);
ArrayList<ArrayList<Integer> > binary_nos =
new ArrayList<ArrayList<Integer> >();
for (int i = l; i <= r; i++) {
ArrayList<Integer> bits =
convertToBinary(i, n);
binary_nos.add(bits);
}
return binary_nos;
}
// Driver code
public static void main(String args[])
{
int L = 2, R = 8;
ArrayList<ArrayList<Integer> > binary_nos =
getAllBinary(L, R);
for (int i = 0; i < binary_nos.size(); i++) {
for (int j = 0; j < binary_nos.get(i).size(); j++) {
System.out.print(binary_nos.get(i).get(j));
}
System.out.println();
}
}
}
// This code is contributed by Samim Hossain Mondal.
Python 3
# Python 3 code to implement the approach
import math
# Function to convert a number to binary
def convertToBinary(num, length):
bits = [0]*(length)
if (num == 0):
return bits
i = length - 1
while (num != 0):
bits[i] = (num % 2)
i -= 1
# Integer division
# gives quotient
num = num // 2
return bits
# Function to convert all numbers
# in range [L, R] to binary of
# same length
def getAllBinary(l, r):
# Length of the binary numbers
n = int(math.ceil(math.log(r+1)/ math.log(2)))
binary_nos = []
for i in range(l, r + 1):
bits = convertToBinary(i, n)
binary_nos.append(bits)
return binary_nos
# Driver code
if __name__ == "__main__":
L = 2
R = 8
binary_nos = getAllBinary(L, R)
for i in range(len(binary_nos)):
for j in range(len(binary_nos[i])):
print(binary_nos[i][j], end="")
print()
# This code is contributed by ukasp.
C#
// C# code to implement the approach
using System;
using System.Collections.Generic;
public class GFG
{
// Function to convert a number to binary
static List<int> convertToBinary(int num, int length)
{
List<int> bits = new List<int>();
int i;
for (i = 0; i < length; i++)
{
bits.Add(0);
}
if (num == 0)
{
return bits;
}
i = length - 1;
while (num != 0)
{
bits[i] = (num % 2);
i = i - 1;
// Integer division
// gives quotient
num = num / 2;
}
return bits;
}
// Function to convert all numbers
// in range [L, R] to binary of
// same length
static List<List<int>> getAllBinary(int l, int r)
{
// Length of the binary numbers
double x = Math.Log(r + 1);
double y = Math.Log(2);
int n = (int)Math.Ceiling(x / y);
List<List<int>> binary_nos = new List<List<int>>();
for (int i = l; i <= r; i++)
{
List<int> bits = convertToBinary(i, n);
binary_nos.Add(bits);
}
return binary_nos;
}
// Driver code
public static void Main()
{
int L = 2, R = 8;
List<List<int>> binary_nos = getAllBinary(L, R);
for (int i = 0; i < binary_nos.Count; i++)
{
for (int j = 0; j < binary_nos[i].Count; j++)
{
Console.Write(binary_nos[i][j]);
}
Console.WriteLine("");
}
}
}
// This code is contributed by Saurabh Jaiswal
java 描述语言
<script>
// JavaScript code to implement the approach
// Function to convert a number to binary
const convertToBinary = (num, length) => {
let bits = new Array(length).fill(0);
if (num == 0) {
return bits;
}
let i = length - 1;
while (num != 0) {
bits[i--] = (num % 2);
// Integer division
// gives quotient
num = parseInt(num / 2);
}
return bits;
}
// Function to convert all numbers
// in range [L, R] to binary of
// same length
const getAllBinary = (l, r) => {
// Length of the binary numbers
let n = Math.ceil(Math.log(r + 1) / Math.log(2));
let binary_nos = [];
for (let i = l; i <= r; i++) {
let bits = convertToBinary(i, n);
binary_nos.push(bits);
}
return binary_nos;
}
// Driver code
let L = 2, R = 8;
let binary_nos = getAllBinary(L, R);
for (let i = 0; i < binary_nos.length;
i++) {
for (let j = 0; j <
binary_nos[i].length; j++)
document.write(binary_nos[i][j]);
document.write("<br/>");
}
// This code is contributed by rakeshsahni
</script>
**Output
0010
0011
0100
0101
0110
0111
1000**
*时间复杂度:* O(N * logR)其中 N =(R–L+1) T3】辅助空间: O(N * logR)
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