检查一个数字的位是否具有递增顺序的连续设置位的计数
原文:https://www . geesforgeks . org/check-bits-number-count-continuous-set-bits-递增-order/
给定一个整数 n > 0，任务是找出在整数计数的位模式中连续的 1 是否在从左向右增加。
例:
Input:19
Output:Yes
Explanation: Bit-pattern of 19 = 10011,
Counts of continuous 1's from left to right
are 1, 2 which are in increasing order.
Input : 183
Output : yes
Explanation: Bit-pattern of 183 = 10110111,
Counts of continuous 1's from left to right
are 1, 2, 3 which are in increasing order.
一个简单的解决方案是将给定数字的二进制表示存储到一个字符串中，然后从左向右遍历并计算连续 1 的数量。对于每次遇到 0，检查连续 1 的先前计数的值与当前值的值，如果先前计数的值大于当前计数的值，则返回假，否则当字符串结束时返回真。
c++
// C++ program to find if bit-pattern
// of a number has increasing value of
// continuous-1 or not.
#include<bits/stdc++.h>
using namespace std;
// Returns true if n has increasing count of
// continuous-1 else false
bool findContinuous1(int n)
{
const int bits = 8*sizeof(int);
// store the bit-pattern of n into
// bit bitset- bp
string bp = bitset <bits>(n).to_string();
// set prev_count = 0 and curr_count = 0.
int prev_count = 0, curr_count = 0;
int i = 0;
while (i < bits)
{
if (bp[i] == '1')
{
// increment current count of continuous-1
curr_count++;
i++;
}
// traverse all continuous-0
else if (bp[i-1] == '0')
{
i++;
curr_count = 0;
continue;
}
// check prev_count and curr_count
// on encounter of first zero after
// continuous-1s
else
{
if (curr_count < prev_count)
return 0;
i++;
prev_count=curr_count;
curr_count = 0;
}
}
// check for last sequence of continuous-1
if (prev_count > curr_count && (curr_count != 0))
return 0;
return 1;
}
// Driver code
int main()
{
int n = 179;
if (findContinuous1(n))
cout << "Yes";
else
cout << "No";
return 0;
}
python 3
# Python3 program to find if bit-pattern
# of a number has increasing value of
# continuous-1 or not.
# Returns true if n has increasing count of
# continuous-1 else false
def findContinuous1(n):
# store the bit-pattern of n into
# bit bitset- bp
bp = list(bin(n))
bits = len(bp)
# set prev_count = 0 and curr_count = 0.
prev_count = 0
curr_count = 0
i = 0
while (i < bits):
if (bp[i] == '1'):
# increment current count of continuous-1
curr_count += 1
i += 1
# traverse all continuous-0
elif (bp[i - 1] == '0'):
i += 1
curr_count = 0
continue
# check prev_count and curr_count
# on encounter of first zero after
# continuous-1s
else:
if (curr_count < prev_count):
return 0
i += 1
prev_count = curr_count
curr_count = 0
# check for last sequence of continuous-1
if (prev_count > curr_count and (curr_count != 0)):
return 0
return 1
# Driver code
n = 179
if (findContinuous1(n)):
print( "Yes")
else:
print( "No")
# This code is contributed by SHUBHAMSINGH10
Javascript
<script>
// JavaScript program to find if bit-pattern
// of a number has increasing value of
// continuous-1 or not.
function dec2bin(dec) {
return (dec >>> 0).toString(2);
}
// Returns true if n has increasing count of
// continuous-1 else false
function findContinuous1(n){
// store the bit-pattern of n into
// bit bitset- bp
let bp = dec2bin(n)
let bits = bp.length
// set prev_count = 0 and curr_count = 0.
let prev_count = 0
let curr_count = 0
let i = 0
while (i < bits){
if (bp[i] == '1'){
// increment current count of continuous-1
curr_count += 1;
i += 1;
}
// traverse all continuous-0
else if (bp[i - 1] == '0'){
i += 1
curr_count = 0
continue
}
// check prev_count and curr_count
// on encounter of first zero after
// continuous-1s
else{
if (curr_count < prev_count)
return 0
i += 1
prev_count = curr_count
curr_count = 0
}
}
// check for last sequence of continuous-1
if (prev_count > curr_count && (curr_count != 0))
return 0
return 1
}
// Driver code
n = 179
if (findContinuous1(n))
document.write( "Yes")
else
document.write( "No")
</script>
输出:
Yes
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