给定两个整数x和n ，任务是搜索长度大于或等于n的第一个连续的1s流(以x的32位二进制表示)，并返回其位置。如果不存在这样的字符串，则返回-1。
例子：
Input: x = 35, n = 2
Output: 31
Binary representation of 35 is 00000000000000000000000000100011 and two consecutive 1’s are present at position 31.
Input: x = 32, n = 3
Output: -1
32 = 00000000000000000000000000100000 in binary and it does not have a sub-string of 3 consecutive 1’s.
方法：使用按位运算来计算编号。的数字中的前导零，然后使用它找到我们需要从中开始搜索连续的1的位置。跳过搜索前导零。
下面是上述方法的实现：
C++
// C++ implementation of above approach
#include
using namespace std;
// Function to count the
// number of leading zeros
int countLeadingZeros(int x)
{
unsigned y;
int n;
n = 32;
y = x >> 16;
if (y != 0) {
n = n - 16;
x = y;
}
y = x >> 8;
if (y != 0) {
n = n - 8;
x = y;
}
y = x >> 4;
if (y != 0) {
n = n - 4;
x = y;
}
y = x >> 2;
if (y != 0) {
n = n - 2;
x = y;
}
y = x >> 1;
if (y != 0)
return n - 2;
return n - x;
}
// Function to find the string
// of n consecutive 1's
int FindStringof1s(unsigned x, int n)
{
int k, p;
// Initialize position to return.
p = 0;
while (x != 0) {
// Skip leading 0's
k = countLeadingZeros(x);
x = x << k;
// Set position after leading 0's
p = p + k;
// Count first group of 1's.
k = countLeadingZeros(~x);
// If length of consecutive 1's
// is greater than or equal to n
if (k >= n)
return p + 1;
// Not enough 1's
// skip over to next group
x = x << k;
// Update the position
p = p + k;
}
// if no string is found
return -1;
}
// Driver code
int main()
{
int x = 35;
int n = 2;
cout << FindStringof1s(x, n);
} Java
// Java implementation of above approach
import java.io.*;
class GFG {
// Function to count the
// number of leading zeros
static int countLeadingZeros(int x)
{
int y;
int n;
n = 32;
y = x >> 16;
if (y != 0) {
n = n - 16;
x = y;
}
y = x >> 8;
if (y != 0) {
n = n - 8;
x = y;
}
y = x >> 4;
if (y != 0) {
n = n - 4;
x = y;
}
y = x >> 2;
if (y != 0) {
n = n - 2;
x = y;
}
y = x >> 1;
if (y != 0)
return n - 2;
return n - x;
}
// Function to find the string
// of n consecutive 1's
static int FindStringof1s(int x, int n)
{
int k, p;
// Initialize position to return.
p = 0;
while (x != 0) {
// Skip leading 0's
k = countLeadingZeros(x);
x = x << k;
// Set position after leading 0's
p = p + k;
// Count first group of 1's.
k = countLeadingZeros(~x);
// If length of consecutive 1's
// is greater than or equal to n
if (k >= n)
return p + 1;
// Not enough 1's
// skip over to next group
x = x << k;
// Update the position
p = p + k;
}
// if no string is found
return -1;
}
// Driver code
public static void main (String[] args) {
int x = 35;
int n = 2;
System.out.println(FindStringof1s(x, n));
}
}C#
// C# implementation of above approach
using System;
public class GFG{
// Function to count the
// number of leading zeros
static int countLeadingZeros(int x)
{
int y;
int n;
n = 32;
y = x >> 16;
if (y != 0) {
n = n - 16;
x = y;
}
y = x >> 8;
if (y != 0) {
n = n - 8;
x = y;
}
y = x >> 4;
if (y != 0) {
n = n - 4;
x = y;
}
y = x >> 2;
if (y != 0) {
n = n - 2;
x = y;
}
y = x >> 1;
if (y != 0)
return n - 2;
return n - x;
}
// Function to find the string
// of n consecutive 1's
static int FindStringof1s(int x, int n)
{
int k, p;
// Initialize position to return.
p = 0;
while (x != 0) {
// Skip leading 0's
k = countLeadingZeros(x);
x = x << k;
// Set position after leading 0's
p = p + k;
// Count first group of 1's.
k = countLeadingZeros(~x);
// If length of consecutive 1's
// is greater than or equal to n
if (k >= n)
return p + 1;
// Not enough 1's
// skip over to next group
x = x << k;
// Update the position
p = p + k;
}
// if no string is found
return -1;
}
// Driver code
static public void Main (){
int x = 35;
int n = 2;
Console.WriteLine (FindStringof1s(x, n));
}
}输出：
31
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