给定一个由 1 和 0 组成的数组，其中先是全 1，然后是全 0。找出 0 的个数。计算给定数组中零的数量。
例子 ：
Input: arr[] = {1, 1, 1, 1, 0, 0}
Output: 2
Input: arr[] = {1, 0, 0, 0, 0}
Output: 4
Input: arr[] = {0, 0, 0}
Output: 3
Input: arr[] = {1, 1, 1, 1}
Output: 0
一个简单的解决方案是遍历输入数组。一旦我们找到一个 0，我们就返回 n——第一个 0 的索引。这里 n 是输入数组中的元素数。该解决方案的时间复杂度为 O(n)。
由于输入数组已排序，我们可以使用二分搜索找到第一次出现的 0。一旦我们有了第一个元素的索引，我们可以将计数返回为 n – 第一个零的索引。
C
// A divide and conquer solution to find count of zeroes in an array
// where all 1s are present before all 0s
#include
/* if 0 is present in arr[] then returns the index of FIRST occurrence
of 0 in arr[low..high], otherwise returns -1 */
int firstZero(int arr[], int low, int high)
{
if (high >= low)
{
// Check if mid element is first 0
int mid = low + (high - low)/2;
if (( mid == 0 || arr[mid-1] == 1) && arr[mid] == 0)
return mid;
if (arr[mid] == 1) // If mid element is not 0
return firstZero(arr, (mid + 1), high);
else // If mid element is 0, but not first 0
return firstZero(arr, low, (mid -1));
}
return -1;
}
// A wrapper over recursive function firstZero()
int countZeroes(int arr[], int n)
{
// Find index of first zero in given array
int first = firstZero(arr, 0, n-1);
// If 0 is not present at all, return 0
if (first == -1)
return 0;
return (n - first);
}
/* Driver program to check above functions */
int main()
{
int arr[] = {1, 1, 1, 0, 0, 0, 0, 0};
int n = sizeof(arr)/sizeof(arr[0]);
printf("Count of zeroes is %d", countZeroes(arr, n));
return 0;
} C++
// A divide and conquer solution to
// find count of zeroes in an array
// where all 1s are present before all 0s
#include
using namespace std;
/* if 0 is present in arr[] then
returns the index of FIRST occurrence
of 0 in arr[low..high], otherwise returns -1 */
int firstZero(int arr[], int low, int high)
{
if (high >= low)
{
// Check if mid element is first 0
int mid = low + (high - low) / 2;
if ((mid == 0 || arr[mid - 1] == 1) &&
arr[mid] == 0)
return mid;
// If mid element is not 0
if (arr[mid] == 1)
return firstZero(arr, (mid + 1), high);
// If mid element is 0, but not first 0
else
return firstZero(arr, low, (mid -1));
}
return -1;
}
// A wrapper over recursive function firstZero()
int countZeroes(int arr[], int n)
{
// Find index of first zero in given array
int first = firstZero(arr, 0, n - 1);
// If 0 is not present at all, return 0
if (first == -1)
return 0;
return (n - first);
}
// Driver Code
int main()
{
int arr[] = {1, 1, 1, 0, 0, 0, 0, 0};
int n = sizeof(arr) / sizeof(arr[0]);
cout << "Count of zeroes is "
<< countZeroes(arr, n);
return 0;
}
// This code is contributed by SoumikMondal Java
// A divide and conquer solution to find count of zeroes in an array
// where all 1s are present before all 0s
class CountZeros
{
/* if 0 is present in arr[] then returns the index of FIRST occurrence
of 0 in arr[low..high], otherwise returns -1 */
int firstZero(int arr[], int low, int high)
{
if (high >= low)
{
// Check if mid element is first 0
int mid = low + (high - low) / 2;
if ((mid == 0 || arr[mid - 1] == 1) && arr[mid] == 0)
return mid;
if (arr[mid] == 1) // If mid element is not 0
return firstZero(arr, (mid + 1), high);
else // If mid element is 0, but not first 0
return firstZero(arr, low, (mid - 1));
}
return -1;
}
// A wrapper over recursive function firstZero()
int countZeroes(int arr[], int n)
{
// Find index of first zero in given array
int first = firstZero(arr, 0, n - 1);
// If 0 is not present at all, return 0
if (first == -1)
return 0;
return (n - first);
}
// Driver program to test above functions
public static void main(String[] args)
{
CountZeros count = new CountZeros();
int arr[] = {1, 1, 1, 0, 0, 0, 0, 0};
int n = arr.length;
System.out.println("Count of zeroes is " + count.countZeroes(arr, n));
}
}Python3
# A divide and conquer solution to
# find count of zeroes in an array
# where all 1s are present before all 0s
# if 0 is present in arr[] then returns
# the index of FIRST occurrence of 0 in
# arr[low..high], otherwise returns -1
def firstZero(arr, low, high):
if (high >= low):
# Check if mid element is first 0
mid = low + int((high - low) / 2)
if (( mid == 0 or arr[mid-1] == 1)
and arr[mid] == 0):
return mid
# If mid element is not 0
if (arr[mid] == 1):
return firstZero(arr, (mid + 1), high)
# If mid element is 0, but not first 0
else:
return firstZero(arr, low, (mid - 1))
return -1
# A wrapper over recursive
# function firstZero()
def countZeroes(arr, n):
# Find index of first zero in given array
first = firstZero(arr, 0, n - 1)
# If 0 is not present at all, return 0
if (first == -1):
return 0
return (n - first)
# Driver Code
arr = [1, 1, 1, 0, 0, 0, 0, 0]
n = len(arr)
print("Count of zeroes is",
countZeroes(arr, n))
# This code is contributed by Smitha Dinesh SemwalC#
// A divide and conquer solution to find
// count of zeroes in an array where all
// 1s are present before all 0s
using System;
class CountZeros
{
/* if 0 is present in arr[] then returns
the index of FIRST occurrence of 0 in
arr[low..high], otherwise returns -1 */
int firstZero(int []arr, int low, int high)
{
if (high >= low)
{
// Check if mid element is first 0
int mid = low + (high - low) / 2;
if ((mid == 0 || arr[mid - 1] == 1) &&
arr[mid] == 0)
return mid;
if (arr[mid] == 1) // If mid element is not 0
return firstZero(arr, (mid + 1), high);
else // If mid element is 0, but not first 0
return firstZero(arr, low, (mid - 1));
}
return -1;
}
// A wrapper over recursive function firstZero()
int countZeroes(int []arr, int n)
{
// Find index of first zero in given array
int first = firstZero(arr, 0, n - 1);
// If 0 is not present at all, return 0
if (first == -1)
return 0;
return (n - first);
}
// Driver program to test above functions
public static void Main()
{
CountZeros count = new CountZeros();
int []arr = {1, 1, 1, 0, 0, 0, 0, 0};
int n = arr.Length;
Console.Write("Count of zeroes is " +
count.countZeroes(arr, n));
}
}
// This code is contributed by nitin mittal.PHP
= $low)
{
// Check if mid element is first 0
$mid = $low + floor(($high - $low)/2);
if (( $mid == 0 || $arr[$mid-1] == 1) &&
$arr[$mid] == 0)
return $mid;
// If mid element is not 0
if ($arr[$mid] == 1)
return firstZero($arr, ($mid + 1), $high);
// If mid element is 0,
// but not first 0
else
return firstZero($arr, $low,
($mid - 1));
}
return -1;
}
// A wrapper over recursive
// function firstZero()
function countZeroes($arr, $n)
{
// Find index of first
// zero in given array
$first = firstZero($arr, 0, $n - 1);
// If 0 is not present
// at all, return 0
if ($first == -1)
return 0;
return ($n - $first);
}
// Driver Code
$arr = array(1, 1, 1, 0, 0, 0, 0, 0);
$n = sizeof($arr);
echo("Count of zeroes is ");
echo(countZeroes($arr, $n));
// This code is contributed by nitin mittal
?>输出：
Count of zeroes is 5 时间复杂度：O(Logn)，其中 n 是 arr[] 中的元素数。
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