用于检查排序数组中的多数元素的Java程序

问题：编写一个函数来查找给定整数 x 在 n 个整数的排序数组中出现的次数是否超过 n/2 次。
基本上，我们需要编写一个函数说 isMajority()，它接受一个数组 (arr[] )、数组的大小 (n) 和一个要搜索的数字 (x) 作为参数，如果 x 是多数元素则返回 true（现在更多大于 n/2 次）。

例子：

Input: arr[] = {1, 2, 3, 3, 3, 3, 10}, x = 3
Output: True (x appears more than n/2 times in the given array)

Input: arr[] = {1, 1, 2, 4, 4, 4, 6, 6}, x = 4
Output: False (x doesn't appear more than n/2 times in the given array)

Input: arr[] = {1, 1, 1, 2, 2}, x = 1
Output: True (x appears more than n/2 times in the given array)

方法 1（使用线性搜索）
线性搜索第一次出现的元素，一旦找到它（让在索引 i 处），检查索引 i + n/2 处的元素。如果元素存在于 i+n/2 处，则返回 1，否则返回 0。

Java

/* Program to check for majority element in a sorted array */
import java.io.*;
  
class Majority {
  
    static boolean isMajority(int arr[], int n, int x)
    {
        int i, last_index = 0;
  
        /* get last index according to n (even or odd) */
        last_index = (n%2==0)? n/2: n/2+1;
  
        /* search for first occurrence of x in arr[]*/
        for (i = 0; i < last_index; i++)
        {
            /* check if x is present and is present more
               than n/2 times */
            if (arr[i] == x && arr[i+n/2] == x)
                return true;
        }
        return false;
    }
  
    /* Driver function to check for above functions*/
    public static void main (String[] args) {
        int arr[] = {1, 2, 3, 4, 4, 4, 4};
        int n = arr.length;
        int x = 4;
        if (isMajority(arr, n, x)==true)
           System.out.println(x+" appears more than "+
                              n/2+" times in arr[]");
        else
           System.out.println(x+" does not appear more than "+
                              n/2+" times in arr[]");
    }
}
/*This article is contributed by Devesh Agrawal*/

Java

import java.util.*;
  
class GFG{
  
static boolean isMajorityElement(int arr[], int n, 
                                 int key)
{
    if (arr[n / 2] == key)
        return true;
    else
        return false;
}
  
// Driver Code
public static void main(String[] args)
{
    int arr[] = { 1, 2, 3, 3, 3, 3, 10 };
    int n = arr.length;
    int x = 3;
      
    if (isMajorityElement(arr, n, x))
        System.out.printf("%d appears more than %d " +
                          "times in arr[]", x, n / 2);
    else
        System.out.printf("%d does not appear more " +
                          "than %d times in " + "arr[]",
                          x, n / 2);
}
}
  
// This code is contributed by aashish1995
Output3 appears more than 3 times in arr[]Time complexity: O(1)Auxiliary Space: O(1)Please refer complete article on Check for Majority Element in a sorted array for more details!

输出：

4 appears more than 3 times in arr[]

时间复杂度： O(n)

方法 2（使用二分搜索）
使用二进制搜索方法查找给定数字的第一次出现。二进制搜索的标准在这里很重要。

Java

/* Java Program to check for majority element in a sorted array */
import java.io.*;
  
class Majority {
  
    /* If x is present in arr[low...high] then returns the index of
        first occurrence of x, otherwise returns -1 */
    static int  _binarySearch(int arr[], int low, int high, int x)
    {
        if (high >= low)
        {
            int mid = (low + high)/2;  /*low + (high - low)/2;*/
  
            /* Check if arr[mid] is the first occurrence of x.
                arr[mid] is first occurrence if x is one of the following
                is true:
                (i)  mid == 0 and arr[mid] == x
                (ii) arr[mid-1] < x and arr[mid] == x
            */
            if ( (mid == 0 || x > arr[mid-1]) && (arr[mid] == x) )
                return mid;
            else if (x > arr[mid])
                return _binarySearch(arr, (mid + 1), high, x);
            else
                return _binarySearch(arr, low, (mid -1), x);
        }
  
        return -1;
    }
  
  
    /* This function returns true if the x is present more than n/2
        times in arr[] of size n */
    static boolean isMajority(int arr[], int n, int x)
    {
        /* Find the index of first occurrence of x in arr[] */
        int i = _binarySearch(arr, 0, n-1, x);
  
        /* If element is not present at all, return false*/
        if (i == -1)
            return false;
  
        /* check if the element is present more than n/2 times */
        if (((i + n/2) <= (n -1)) && arr[i + n/2] == x)
            return true;
        else
            return false;
    }
  
    /*Driver function to check for above functions*/
    public static void main (String[] args)  {
  
        int arr[] = {1, 2, 3, 3, 3, 3, 10};
        int n = arr.length;
        int x = 3;
        if (isMajority(arr, n, x)==true)
            System.out.println(x + " appears more than "+
                              n/2 + " times in arr[]");
        else
            System.out.println(x + " does not appear more than " +
                              n/2 + " times in arr[]");
    }
}
/*This code is contributed by Devesh Agrawal*/

输出：

3 appears more than 3 times in arr[]

时间复杂度： O(Logn)
算法范式：分而治之

方法3：如果已经给定数组已排序并且存在多数元素，则检查特定元素是否与检查数组的中间元素是否是我们要检查的数字一样简单。

由于多数元素在数组中出现超过 n/2 次，因此它将始终是中间元素。我们可以使用这个逻辑来检查给定的数字是否是多数元素。

Java

import java.util.*;
  
class GFG{
  
static boolean isMajorityElement(int arr[], int n, 
                                 int key)
{
    if (arr[n / 2] == key)
        return true;
    else
        return false;
}
  
// Driver Code
public static void main(String[] args)
{
    int arr[] = { 1, 2, 3, 3, 3, 3, 10 };
    int n = arr.length;
    int x = 3;
      
    if (isMajorityElement(arr, n, x))
        System.out.printf("%d appears more than %d " +
                          "times in arr[]", x, n / 2);
    else
        System.out.printf("%d does not appear more " +
                          "than %d times in " + "arr[]",
                          x, n / 2);
}
}
  
// This code is contributed by aashish1995

输出

3 appears more than 3 times in arr[]

时间复杂度：O(1)
辅助空间： O(1)

有关详细信息，请参阅有关在排序数组中检查多数元素的完整文章！

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