Swift-魔术索引

题目:在数组A[0..n-1]中，有所谓的魔术索引，满足条件A[i]=i。给定一个升序数组，元素值各不相同，编写一个方法，判断在数组A中是否存在魔术索引.

数组中不存在重复数据可以通过二分法来实现:

func magicFast(arr:inout [Int],start:Int,end:Int) -> Int {
        if arr.count == 0 || start < 0 || end >= arr.count || end < start {
            return -1
        }
        
        let mid:Int = (start + end) / 2
        
        let midValue:Int = arr[mid]
        
        if mid == midValue {
            return mid
        } else if midValue > mid {
            return magicFast(arr: &arr, start: start, end: mid - 1)
        } else {
             return magicFast(arr: &arr, start: mid + 1, end: end)
        }
    }

数据中存在重复数据:

   func magicFast2(arr:inout [Int],start:Int,end:Int) -> Int {
        if arr.count == 0 || start < 0 || end >= arr.count || end < start {
            return -1
        }
        
        let mid:Int = (start + end) / 2
        
        let midValue:Int = arr[mid]
        
        
        if mid == midValue {
            return mid
        }
        
        let leftIndex:Int = min(mid - 1, midValue)
        let left:Int = magicFast2(arr: &arr, start: start, end: leftIndex)
        if left >= 0 {
            return left
        }
        
        let rightIndex:Int = max(mid + 1, midValue)
        let right:Int = magicFast2(arr: &arr, start: rightIndex, end: end)
        return right
    }

测试代码:

var magicArr:[Int] = [-40, -20, -1, 1, 2, 3, 5, 7, 9, 12, 13]
var magicIndex:Int = recursion.magicFast(arr: &magicArr, start: 0, end: magicArr.count - 1)
print("FlyElephant---\(magicArr)中间值:\(magicIndex)")

magicArr = [-10, -5, 2, 2, 2, 3, 4, 7, 9, 12, 13]
magicIndex = recursion.magicFast2(arr: &magicArr, start: 0, end: magicArr.count - 1)
print("FlyElephant---\(magicArr)中间值:\(magicIndex)")

FlyElephant.png