给定n个不同整数的数组arr [] 。元素按升序顺序放置，而缺少一个元素。任务是找到缺少的元素。
例子：
Input: arr[] = {1, 2, 4, 5, 6, 7, 8, 9}
Output: 3
Input: arr[] = {-4, -3, -1, 0, 1, 2}
Output: -2
Input: arr[] = {1, 2, 3, 4}
Output: -1
No element is missing.
原则：
- Look for inconsistency: Ideally, the difference between any element and its index must be arr[0] for every element.
Example,
A[] = {1, 2, 3, 4, 5} -> Consistent
B[] = {101, 102, 103, 104} -> Consistent
C[] = {1, 2, 4, 5, 6} -> Inconsistent as C[2] – 2 != C[0] i.e. 4 – 2 != 1 - Finding inconsistency helps to scan only half of the array each time in O(logN).
算法
- Find middle element and check if it’s consistent.
- If middle element is consistent, then check if the difference between middle element and its next element is greater than 1 i.e. check if arr[mid + 1] – arr[mid] > 1
- If yes, then arr[mid] + 1 is the missing element.
- If not, then we have to scan the right half array from the middle element and jump to step-1.
- If middle element is inconsistent, then check if the difference between middle element and its previous element is greater than 1 i.e. check if arr[mid] – arr[mid – 1] > 1
- If yes, then arr[mid] – 1 is the missing element.
- If not, then we have to scan the left half array from the middle element and jump to step-1.
下面是上述方法的实现：
C++
// CPP implementation of the approach
#include
using namespace std;
// Function to return the missing element
int findMissing(int arr[], int n)
{
int l = 0, h = n - 1;
int mid;
while (h > l)
{
mid = l + (h - l) / 2;
// Check if middle element is consistent
if (arr[mid] - mid == arr[0])
{
// No inconsistency till middle elements
// When missing element is just after
// the middle element
if (arr[mid + 1] - arr[mid] > 1)
return arr[mid] + 1;
else
{
// Move right
l = mid + 1;
}
}
else
{
// Inconsistency found
// When missing element is just before
// the middle element
if (arr[mid] - arr[mid - 1] > 1)
return arr[mid] - 1;
else
{
// Move left
h = mid - 1;
}
}
}
// No missing element found
return -1;
}
// Driver code
int main()
{
int arr[] = { -9, -8, -7, -5, -4, -3, -2, -1, 0 };
int n = sizeof(arr)/sizeof(arr[0]);
cout << (findMissing(arr, n));
}
// This code iscontributed by
// Surendra_Gangwar Java
// Java implementation of the approach
class GFG {
// Function to return the missing element
public static int findMissing(int arr[], int n)
{
int l = 0, h = n - 1;
int mid;
while (h > l) {
mid = l + (h - l) / 2;
// Check if middle element is consistent
if (arr[mid] - mid == arr[0]) {
// No inconsistency till middle elements
// When missing element is just after
// the middle element
if (arr[mid + 1] - arr[mid] > 1)
return arr[mid] + 1;
else {
// Move right
l = mid + 1;
}
}
else {
// Inconsistency found
// When missing element is just before
// the middle element
if (arr[mid] - arr[mid - 1] > 1)
return arr[mid] - 1;
else {
// Move left
h = mid - 1;
}
}
}
// No missing element found
return -1;
}
// Driver code
public static void main(String args[])
{
int arr[] = { -9, -8, -7, -5, -4, -3, -2, -1, 0 };
int n = arr.length;
System.out.print(findMissing(arr, n));
}
}Python3
# Python implementation of the approach
# Function to return the missing element
def findMissing(arr, n):
l, h = 0, n - 1
mid = 0
while (h > l):
mid = l + (h - l) // 2
# Check if middle element is consistent
if (arr[mid] - mid == arr[0]):
# No inconsistency till middle elements
# When missing element is just after
# the middle element
if (arr[mid + 1] - arr[mid] > 1):
return arr[mid] + 1
else:
# Move right
l = mid + 1
else:
# Inconsistency found
# When missing element is just before
# the middle element
if (arr[mid] - arr[mid - 1] > 1):
return arr[mid] - 1
else:
# Move left
h = mid - 1
# No missing element found
return -1
# Driver code
arr = [-9, -8, -7, -5, -4, -3, -2, -1, 0 ]
n = len(arr)
print(findMissing(arr, n))
# This code is contributed
# by mohit kumarC#
// C# implementation of the approach
using System;
class GFG
{
// Function to return the missing element
public static int findMissing(int[] arr, int n)
{
int l = 0, h = n - 1;
int mid;
while (h > l)
{
mid = l + (h - l) / 2;
// Check if middle element is consistent
if (arr[mid] - mid == arr[0])
{
// No inconsistency till middle elements
// When missing element is just after
// the middle element
if (arr[mid + 1] - arr[mid] > 1)
return arr[mid] + 1;
else
{
// Move right
l = mid + 1;
}
}
else
{
// Inconsistency found
// When missing element is just before
// the middle element
if (arr[mid] - arr[mid - 1] > 1)
return arr[mid] - 1;
else
{
// Move left
h = mid - 1;
}
}
}
// No missing element found
return -1;
}
// Driver code
public static void Main()
{
int[] arr = { -9, -8, -7, -5, -4, -3, -2, -1, 0 };
int n = arr.Length;
Console.WriteLine(findMissing(arr, n));
}
}
// This code is contributed by Code_MechPHP
$l)
{
$mid = floor($l + ($h - $l) / 2);
// Check if middle element is consistent
if ($arr[$mid] - $mid == $arr[0])
{
// No inconsistency till middle elements
// When missing element is just after
// the middle element
if ($arr[$mid + 1] - $arr[$mid] > 1)
return $arr[$mid] + 1;
else
{
// Move right
$l = $mid + 1;
}
}
else
{
// Inconsistency found
// When missing element is just before
// the middle element
if ($arr[$mid] - $arr[$mid - 1] > 1)
return $arr[$mid] - 1;
else
{
// Move left
$h = $mid - 1;
}
}
}
// No missing element found
return -1;
}
// Driver code
$arr = array( -9, -8, -7, -5, -
4, -3, -2, -1, 0 );
$n = count($arr);
echo findMissing($arr, $n);
// This code is contributed by Ryuga
?>输出：
-6