Glede na številko n Naloga je najti dolžino najdaljšega zaporedja 1s serije v svoji binarni predstavitvi.
Primeri:
Vnos: n = 14
Izhod: 3
Pojasnilo: Binarna predstavitev 14 je 111 0.
Vnos: n = 222
Izhod: 4
Pojasnilo: Binarna predstavitev števila 222 je 110 1111 0.
Kazalo vsebine
- [Naivni pristop] Iterativni čas O(1) in prostor O(1)
- [Učinkovit pristop] Uporaba bitne manipulacije O(1) Čas in O(1) Prostor
- [Drug pristop] Uporaba pretvorbe nizov
[Naivni pristop] Iterativni čas O(1) in prostor O(1)
C++#include
public class GFG { static int maxConsecutiveOne(int n) { int count = 0; int maxi = 0; // traverse and check if bit set increment the count for (int i = 0; i < 32; i++) { if ((n & (1 << i)) != 0) { count++; } else { maxi = Math.max(maxi count); count = 0; } } return maxi; } public static void main(String[] args) { int n = 14; System.out.println(maxConsecutiveOne(n)); } }
def maxConsecutiveOne(n): count = 0 maxi = 0 # traverse and check if bit set increment the count for i in range(32): if n & (1 << i): count += 1 else: maxi = max(maxi count) count = 0 return maxi if __name__ == '__main__': n = 14 print(maxConsecutiveOne(n))
using System; class GFG { static int MaxConsecutiveOne(int n) { int count = 0; int maxi = 0; // traverse and check if bit set increment the count for (int i = 0; i < 32; i++) { if ((n & (1 << i)) != 0) { count++; } else { maxi = Math.Max(maxi count); count = 0; } } return maxi; } static void Main() { int n = 14; Console.WriteLine(MaxConsecutiveOne(n)); } }
function maxConsecutiveOne(n) { let count = 0; let maxi = 0; // traverse and check if bit set increment the count for (let i = 0; i < 32; i++) { if (n & (1 << i)) { count++; } else { maxi = Math.max(maxi count); count = 0; } } return maxi; } // Driver code let n = 14; console.log(maxConsecutiveOne(n));
Izhod
3
[Učinkovit pristop] Uporaba bitne manipulacije O(1) Čas in O(1) Prostor
Ideja temelji na konceptu, da je IN bitnega zaporedja z a levo premaknjeno za 1 sama različica učinkovito odstrani zaostanek 1 iz vsakega zaporednega zaporedja 1s .
Torej operacija n = (n & (n<< 1)) zmanjša dolžino vsakega zaporedja 1s z eno v binarni predstavitvi n . Če to operacijo izvajamo v zanki, dobimo n = 0. Število ponovitev, potrebnih za dosego je pravzaprav dolžina najdaljšega zaporednega zaporedja 1s .
Ilustracija:
Za izvedbo zgornjega pristopa sledite spodnjim korakom:
- Ustvarite spremenljivko count, inicializirano z vrednostjo .
- Zaženite zanko medtem do n ni 0.
- V vsaki ponovitvi izvedite operacijo n = (n & (n<< 1))
- Povečaj štetje za eno.
- Povratno število
#include
class GFG { private static int maxConsecutiveOnes(int x) { // Initialize result int count = 0; // Count the number of iterations to // reach x = 0. while (x!=0) { // This operation reduces length // of every sequence of 1s by one. x = (x & (x << 1)); count++; } return count; } public static void main(String strings[]) { System.out.println(maxConsecutiveOnes(14)); } }
def maxConsecutiveOnes(x): # Initialize result count = 0 # Count the number of iterations to # reach x = 0. while (x!=0): # This operation reduces length # of every sequence of 1s by one. x = (x & (x << 1)) count=count+1 return count if __name__ == '__main__': print(maxConsecutiveOnes(14)) # by Anant Agarwal.
using System; class GFG { // Function to find length of the // longest consecutive 1s in binary // representation of a number private static int maxConsecutiveOnes(int x) { // Initialize result int count = 0; // Count the number of iterations // to reach x = 0. while (x != 0) { // This operation reduces length // of every sequence of 1s by one. x = (x & (x << 1)); count++; } return count; } // Driver code public static void Main() { Console.WriteLine(maxConsecutiveOnes(14)); } } // This code is contributed by Nitin Mittal.
function maxConsecutiveOnes(x) { // Initialize result let count = 0; // Count the number of iterations to reach x = 0 while (x !== 0) { // This operation reduces length of // every sequence of 1s by one x = (x & (x << 1)); count++; } return count; } // Driver code console.log(maxConsecutiveOnes(14));
// PHP program to find length function maxConsecutiveOnes($n) { // Initialize result $count = 0; // Count the number of // iterations to reach x = 0. while ($n != 0) { // This operation reduces // length of every sequence // of 1s by one. $n = ($n & ($n << 1)); $count++; } return $count; } echo maxConsecutiveOnes(14) 'n'; ?> Izhod
3
Časovna zapletenost: O(1)
Pomožni prostor: O(1)
[Drug pristop] Uporaba pretvorbe nizov
Dve spremenljivki max_len in cur_len inicializiramo na 0. Nato ponovimo vsak bit celega števila n. Če je najmanj pomemben bit (LSB) 1, povečamo cur_len, da preštejemo trenutni zagon zaporednih 1 s. Če je LSB 0, prekine trenutno zaporedje, tako da posodobimo max_len, če je cur_len večji, in ponastavimo cur_len na 0. Po preverjanju vsakega bita premaknemo n v desno za 1, da se premaknemo na naslednji bit. Končno po koncu zanke izvedemo zadnjo posodobitev max_len, če je končni cur_len večji, in vrnemo max_len kot dolžino najdaljšega zaporedja zaporednih 1 s.
C++#include
import java.util.*; public class Main { static int maxConsecutiveOnes(int n) { String binary = String.format('%32s' Integer.toBinaryString(n)).replace(' ' '0'); int count = 0; int maxCount = 0; // Loop through the binary string to // find the longest consecutive 1s for (int i = 0; i < binary.length(); i++) { if (binary.charAt(i) == '1') { count++; if (count > maxCount) { maxCount = count; } } else { count = 0; } } // Return the result return maxCount; } public static void main(String[] args) { int n = 14; System.out.println(maxConsecutiveOnes(n)); } }
def maxConsecutiveOnes(n): binary = format(n '032b') count = 0 maxCount = 0 # Loop through the binary string to # find the longest consecutive 1s for bit in binary: if bit == '1': count += 1 if count > maxCount: maxCount = count else: count = 0 # Return the result return maxCount if __name__ == '__main__': n = 14 print(maxConsecutiveOnes(n))
using System; class GFG { static int MaxConsecutiveOnes(int n) { string binary = Convert.ToString(n 2).PadLeft(32 '0'); int count = 0; int maxCount = 0; // Loop through the binary string to // find the longest consecutive 1s foreach (char bit in binary) { if (bit == '1') { count++; if (count > maxCount) maxCount = count; } else { count = 0; } } // Return the result return maxCount; } static void Main() { int n = 14; Console.WriteLine(MaxConsecutiveOnes(n)); } }
function maxConsecutiveOnes(n) { let binary = n.toString(2).padStart(32 '0'); let count = 0; let maxCount = 0; // Loop through the binary string to // find the longest consecutive 1s for (let i = 0; i < binary.length; i++) { if (binary[i] === '1') { count++; if (count > maxCount) { maxCount = count; } } else { count = 0; } } // Return the result return maxCount; } // Driver code let n = 14; console.log(maxConsecutiveOnes(n));
Izhod
3