我们给了n个整数的只读数组。在线性时间和恒定的额外空间中找到数组中出现次数超过n / 3次的任何元素。如果不存在这样的元素，则返回-1。
例子：
Input : [10, 10, 20, 30, 10, 10]
Output : 10
10 occurs 4 times which is more than 6/3.
Input : [20, 30, 10, 10, 5, 4, 20, 1, 2]
Output : -1这个想法是基于摩尔的投票算法。我们首先找到两个候选人。然后，我们检查这两个候选人中是否有任何一个真正占多数。以下是上述方法的解决方案。
C++
// CPP program to find if any element appears
// more than n/3.
#include
using namespace std;
int appearsNBy3(int arr[], int n)
{
int count1 = 0, count2 = 0;
int first=INT_MAX , second=INT_MAX ;
for (int i = 0; i < n; i++) {
// if this element is previously seen,
// increment count1.
if (first == arr[i])
count1++;
// if this element is previously seen,
// increment count2.
else if (second == arr[i])
count2++;
else if (count1 == 0) {
count1++;
first = arr[i];
}
else if (count2 == 0) {
count2++;
second = arr[i];
}
// if current element is different from
// both the previously seen variables,
// decrement both the counts.
else {
count1--;
count2--;
}
}
count1 = 0;
count2 = 0;
// Again traverse the array and find the
// actual counts.
for (int i = 0; i < n; i++) {
if (arr[i] == first)
count1++;
else if (arr[i] == second)
count2++;
}
if (count1 > n / 3)
return first;
if (count2 > n / 3)
return second;
return -1;
}
int main()
{
int arr[] = { 1, 2, 3, 1, 1 };
int n = sizeof(arr) / sizeof(arr[0]);
cout << appearsNBy3(arr, n) << endl;
return 0;
} Java
// Java program to find if any element appears
// more than n/3.
class GFG {
static int appearsNBy3(int arr[], int n)
{
int count1 = 0, count2 = 0;
// take the integers as the maximum
// value of integer hoping the integer
// would not be present in the array
int first = Integer.MIN_VALUE;;
int second = Integer.MAX_VALUE;
for (int i = 0; i < n; i++) {
// if this element is previously
// seen, increment count1.
if (first == arr[i])
count1++;
// if this element is previously
// seen, increment count2.
else if (second == arr[i])
count2++;
else if (count1 == 0) {
count1++;
first = arr[i];
}
else if (count2 == 0) {
count2++;
second = arr[i];
}
// if current element is different
// from both the previously seen
// variables, decrement both the
// counts.
else {
count1--;
count2--;
}
}
count1 = 0;
count2 = 0;
// Again traverse the array and
// find the actual counts.
for (int i = 0; i < n; i++) {
if (arr[i] == first)
count1++;
else if (arr[i] == second)
count2++;
}
if (count1 > n / 3)
return first;
if (count2 > n / 3)
return second;
return -1;
}
// Driver code
public static void main(String args[])
{
int arr[] = { 1, 2, 3, 1, 1 };
int n = arr.length;
System.out.println(appearsNBy3(arr, n));
}
}
// This code is contributed by Arnab KunduPython 3
# Python 3 program to find if
# any element appears more than
# n/3.
import sys
def appearsNBy3(arr, n):
count1 = 0
count2 = 0
first = sys.maxsize
second = sys.maxsize
for i in range(0, n):
# if this element is
# previously seen,
# increment count1.
if (first == arr[i]):
count1 += 1
# if this element is
# previously seen,
# increment count2.
elif (second == arr[i]):
count2 += 1
elif (count1 == 0):
count1 += 1
first = arr[i]
elif (count2 == 0):
count2 += 1
second = arr[i]
# if current element is
# different from both
# the previously seen
# variables, decrement
# both the counts.
else:
count1 -= 1
count2 -= 1
count1 = 0
count2 = 0
# Again traverse the array
# and find the actual counts.
for i in range(0, n):
if (arr[i] == first):
count1 += 1
elif (arr[i] == second):
count2 += 1
if (count1 > n / 3):
return first
if (count2 > n / 3):
return second
return -1
# Driver code
arr = [1, 2, 3, 1, 1 ]
n = len(arr)
print(appearsNBy3(arr, n))
# This code is contributed by
# SmithaC#
// C# program to find if any element appears
// more than n/3.
using System;
public class GFG {
static int appearsNBy3(int []arr, int n)
{
int count1 = 0, count2 = 0;
// take the integers as the maximum
// value of integer hoping the integer
// would not be present in the array
int first = int.MaxValue;
int second = int.MaxValue;
for (int i = 1; i < n; i++) {
// if this element is previously
// seen, increment count1.
if (first == arr[i])
count1++;
// if this element is previously
// seen, increment count2.
else if (second == arr[i])
count2++;
else if (count1 == 0) {
count1++;
first = arr[i];
}
else if (count2 == 0) {
count2++;
second = arr[i];
}
// if current element is different
// from both the previously seen
// variables, decrement both the
// counts.
else {
count1--;
count2--;
}
}
count1 = 0;
count2 = 0;
// Again traverse the array and
// find the actual counts.
for (int i = 0; i < n; i++) {
if (arr[i] == first)
count1++;
else if (arr[i] == second)
count2++;
}
if (count1 > n / 3)
return first;
if (count2 > n / 3)
return second;
return -1;
}
// Driver code
static public void Main(String []args)
{
int []arr = { 1, 2, 3, 1, 1 };
int n = arr.Length;
Console.WriteLine(appearsNBy3(arr, n));
}
}
// This code is contributed by Arnab KunduPHP
$n / 3)
return $first;
if ($count2 > $n / 3)
return $second;
return -1;
}
// Driver code
$arr = array( 1, 2, 3, 1, 1 );
$n = count($arr);
echo appearsNBy3($arr, $n) ;
// This code is contributed by anuj_67.
?>输出：
1复杂度分析：
- 时间复杂度： O(n)
该算法的第一遍对构成O(n)的数组进行完整遍历，并在检查count1和count2是否大于floor(n / 3)次时消耗了另一个O(n)。 - 空间复杂度：O(1)
由于不需要额外的空间，因此空间复杂度是恒定的