Javascript程序计算排序二进制数组中的1

给定一个按非递增顺序排序的二进制数组，计算其中 1 的个数。

例子：

Input: arr[] = {1, 1, 0, 0, 0, 0, 0}
Output: 2

Input: arr[] = {1, 1, 1, 1, 1, 1, 1}
Output: 7

Input: arr[] = {0, 0, 0, 0, 0, 0, 0}
Output: 0

一个简单的解决方案是线性遍历数组。简单解的时间复杂度为 O(n)。我们可以使用二分搜索在 O(Logn) 时间内找到计数。这个想法是使用二分搜索查找 1 的最后一次出现。一旦我们找到最后一次出现的索引，我们就返回 index + 1 作为计数。
以下是上述思想的实现。

Javascript

  
// Javascript program to count one's in a boolean array
  
/* Returns counts of 1's in arr[low..high].  The array is
   assumed to be sorted in non-increasing order */
function countOnes( arr, low, high)
{
  if (high >= low)
  {
    // get the middle index
    let mid = Math.trunc(low + (high - low)/2);
  
    // check if the element at middle index is last 1
    if ( (mid == high || arr[mid+1] == 0) && (arr[mid] == 1))
      return mid+1;
  
    // If element is not last 1, recur for right side
    if (arr[mid] == 1)
      return countOnes(arr, (mid + 1), high);
  
    // else recur for left side
    return countOnes(arr, low, (mid -1));
  }
  return 0;
}
    // Driver program 
      
   let arr = [ 1, 1, 1, 1, 0, 0, 0 ];
   let n = arr.length;
   document.write("Count of 1's in given array is " + 
                    countOnes(arr, 0, n-1));
      

输出

Count of 1's in given array is 4

上述解决方案的时间复杂度为 O(Logn)

空间复杂度 o(log n) (函数调用栈)

迭代解决方案的相同方法是

Javascript

/* Returns counts of 1's in arr[low..high].  The array is
   assumed to be sorted in non-increasing order */
  
function countOnes(arr,n)
{
    let ans;
    let low = 0, high = n - 1;
    while (low <= high) { // get the middle index
        let mid = Math.floor((low + high) / 2);
   
        // else recur for left side
        if (arr[mid] < 1)
            high = mid - 1;
        // If element is not last 1, recur for right side
        else if (arr[mid] > 1)
            low = mid + 1;
        else
        // check if the element at middle index is last 1
        {
            if (mid == n - 1 || arr[mid + 1] != 1)
                return mid + 1;
            else
                low = mid + 1;
        }
    }
}
  
let arr=[ 1, 1, 1, 1, 0, 0, 0];
let n = arr.length;
document.write( "Count of 1's in given array is "+ countOnes(arr, n));
  
  
// This code is contributed by unknown2108

输出

Count of 1's in given array is 4

上述解决方案的时间复杂度为 O(Logn)

空间复杂度为 O(1)

有关详细信息，请参阅有关排序二进制数组中计数 1 的完整文章！