題目：統計一個數字在排序數組中出現的次數。例如輸入排序數組{1,2,3,3,3,3,4,5}和數字3，由于3在這個數組中出現了4次，因此輸出4.
最簡單的是一次遍歷，但是更高效的方法是二分查找，需要略微改變.
核心代碼:
<pre><code>`
func searchNumberOfK(arr:[Int],num:Int) -> Int? {
if arr.count == 0 {
return nil
}

    let first:Int? = searchFirstOfK(arr: arr, num: num)
    let last:Int? = searchLastOfK(arr: arr, num: num)
    if first == nil || last == nil {
        return nil
    } else {
        let count:Int = last! - first! + 1
        return count
    }
}

func searchFirstOfK(arr:[Int],num:Int) -> Int? {
    var low:Int = 0
    var high:Int = arr.count - 1
    while low <= high {
        let mid:Int = (low + high)/2
        if arr[mid] > num {
            high = mid - 1
        } else if arr[mid] < num {
            low = mid + 1
        } else {
            if  (mid > 0 && arr[mid-1] != num) || mid == 0 {
                return mid
            } else {
                high = mid - 1
            }
        }
    }
    return nil
}

func searchLastOfK(arr:[Int],num:Int) -> Int? {
    var low:Int = 0
    var high:Int = arr.count - 1
    while low <= high {
        let mid:Int = (low + high)/2
        if arr[mid] > num {
            high = mid - 1
        } else if arr[mid] < num {
            low = mid + 1
        } else {
            if  (mid < arr.count - 1 && arr[mid+1] != num) || mid == arr.count - 1 {
                return mid
            } else {
                low = mid + 1
            }
        }
    }
    return nil
}`</code></pre>

測試代碼:
<pre><code>`

var sortArr:[Int] = [1,2,3,3,3,3,4,5]
var sortTarget:Int = 1
var sortCount:Int? = binarySearch.searchNumberOfK(arr: sortArr, num: sortTarget)
if sortCount == nil {
print("FlyElephant-(sortArr)不存在數字-(sortTarget)")
} else {
print("FlyElephant-(sortArr)中數字-(sortTarget)的次數--(sortCount!)")`</code></pre>