Javascript程序检查排序数组中的多数元素

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。

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。

Javascript

    // Javascript Program to check for majority
    // element in a sorted array */
      
    // If x is present in arr[low...high]
    // then returns the index of first
    // occurrence of x, otherwise returns -1
    function _binarySearch(arr, low, high, x)
    {
        if (high >= low) {
            let mid = parseInt((low + high) / 2, 10);
            //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
    function isMajority(arr, n, x)
    {
           
        // Find the index of first occurrence
        // of x in arr[]
        let 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 + parseInt(n / 2, 10)) <= (n - 1)) && arr[i + parseInt(n / 2, 10)] == x)
            return true;
        else
            return false;
    }
      
    let arr = [ 1, 2, 3, 3, 3, 3, 10 ];
    let n = arr.length;
    let x = 3;
    if (isMajority(arr, n, x) == true)
      document.write(x + " appears more than " + parseInt(n / 2, 10) + " times in arr[]");
    else
      document.write(x + " does not appear more than " + parseInt(n / 2, 10) + " times in arr[]");
      

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！