一个数字的每一个数字最接近的 2 的小次方
原文:https://www . geeksforgeeks . org/number 的每一位都是最近的 2 的小次方/
给定一个整数数，这个数的每个数字的任务是找到 2 的最高幂，不超过那个数字。
示例:
输入: num = 4317 输出: 4214 说明: 2≤4 的最高幂是 4。 2≤3 的最高幂是 2。 2≤1 的最高幂为 1。 2≤7 的最高幂为 4。
输入:num = 8015 T3】输出: 8014
方法:按照以下步骤解决问题:
- 将数字转换为其等价字符串。
- 穿越弦。
- 如果数字是‘0’，打印 0 。
- 否则，对于每一个数字 x ，计算 2 (日志T52T8(x)T11】。
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 find the nearest power of
// two for every digit of a given number
void highestPowerOfTwo(int num)
{
// Converting number to string
string s = to_string(num);
// Traverse the array
for (int i = 0; i < (int)s.size();
i++) {
if (s[i] == '0') {
cout << "0";
continue;
}
// Calculate log base 2
// of the current digit s[i]
int lg = log2(int(s[i]) - 48);
// Highest power of 2 <= s[i]
int p = pow(2, lg);
// ASCII conversion
cout << char(p + 48);
}
}
// Driver Code
int main()
{
int num = 4317;
highestPowerOfTwo(num);
return 0;
}
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.util.*;
class GFG
{
// Function to find the nearest power of
// two for every digit of a given number
static void highestPowerOfTwo(int num)
{
// Converting number to string
String s = Integer.toString(num);
// Traverse the array
for (int i = 0; i < (int)s.length(); i++)
{
if (s.charAt(i) == '0')
{
System.out.print("0");
continue;
}
// Calculate log base 2
// of the current digit s[i]
int lg
= (int)(Math.log(s.charAt(i) - '0') / Math.log(2));
// Highest power of 2 <= s[i]
int p = (int)Math.pow(2, lg);
// ASCII conversion
System.out.print((char)(p + 48));
}
}
// Driver Code
public static void main(String args[])
{
int num = 4317;
highestPowerOfTwo(num);
}
}
// This code is contributed by susmitakundugoaldanga.
Python 3
# Python 3 program for the above approach
import math
# Function to find the nearest power of
# two for every digit of a given number
def highestPowerOfTwo(num) :
# Converting number to string
s = str(num)
# Traverse the array
for i in range(len(s)):
if (s[i] == '0') :
print("0")
continue
# Calculate log base 2
# of the current digit s[i]
lg = int(math.log2(ord(s[i]) - 48))
# Highest power of 2 <= s[i]
p = pow(2, lg)
# ASCII conversion
print(chr(p + 48), end = "")
# Driver Code
num = 4317
highestPowerOfTwo(num)
# This code is contributed by code_hunt.
C
// C# program to implement
// the above approach
using System;
class GFG
{
// Function to find the nearest power of
// two for every digit of a given number
static void highestPowerOfTwo(int num)
{
// Converting number to string
String s = num.ToString();
// Traverse the array
for (int i = 0; i < (int)s.Length; i++)
{
if (s[i] == '0')
{
Console.Write("0");
continue;
}
// Calculate log base 2
// of the current digit s[i]
int lg
= (int)(Math.Log(s[i] - '0') / Math.Log(2));
// Highest power of 2 <= s[i]
int p = (int)Math.Pow(2, lg);
// ASCII conversion
Console.Write((char)(p + 48));
}
}
// Driver Code
public static void Main()
{
int num = 4317;
highestPowerOfTwo(num);
}
}
// This code is contributed by subhammahato348.
java 描述语言
<script>
// JavaScript program to implement
// the above approach
// Function to find the nearest power of
// two for every digit of a given number
function highestPowerOfTwo(num) {
// Converting number to string
var s = num.toString();
// Traverse the array
for (var i = 0; i < s.length; i++) {
if (s[i] === "0") {
document.write("0");
continue;
}
// Calculate log base 2
// of the current digit s[i]
var lg = parseInt(Math.log2(s[i].charCodeAt(0) - 48));
// Highest power of 2 <= s[i]
var p = Math.pow(2, lg);
// ASCII conversion
document.write(String.fromCharCode(p + 48));
}
}
// Driver Code
var num = 4317;
highestPowerOfTwo(num);
</script>
Output:
4214
时间复杂度: O(logN) 辅助空间: O(1)
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