[Swift]LeetCode69. x 的平方根 | Sqrt(x)

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Implement int sqrt(int x).

Compute and return the square root of x, where x is guaranteed to be a non-negative integer.

Since the return type is an integer, the decimal digits are truncated and only the integer part of the result is returned.

Example 1:

Input: 4
Output: 2

Example 2:

Input: 8
Output: 2
Explanation: The square root of 8 is 2.82842..., and since 
             the decimal part is truncated, 2 is returned.

---

实现 int sqrt(int x) 函数。

计算并返回 x 的平方根，其中 x 是非负整数。

由于返回类型是整数，结果只保留整数的部分，小数部分将被舍去。

示例 1:

输入: 4
输出: 2

示例 2:

输入: 8
输出: 2
说明: 8 的平方根是 2.82842..., 
     由于返回类型是整数，小数部分将被舍去。

---

二分法用除法（20ms）

class Solution {
    func mySqrt(_ x: Int) -> Int {
        //二分法
        var low:Int = 1
        var high:Int = x
        while(low <= high)
        {
            var mid = (low+high)/2
            //乘法会导致溢出，应写成除法
            if mid == x/mid
            {
                return mid
            }
            else
            {
                if mid < x/mid
                {
                    low = mid + 1
                }
                else
                {
                    high = mid - 1
                }
            }
        }
        return low-1  
    }
}

---

二分法用乘法（24ms），可能导致溢出

class Solution {
    func mySqrt(_ x: Int) -> Int {
        if x < 2 { return x }
        
        var low = 1
        var high = x / 2
        
        while (low <= high) {
            let mid = low + (high - low) / 2
            if mid * mid <= x && (mid+1) * (mid+1) > x {
                return mid
            } else if mid * mid <= x && (mid+1) * (mid+1) <= x {
                low = mid + 1
            } else {
                high = mid - 1
            }
        }
        
        return -1
    }
}

---

牛顿迭代法：20ms

class Solution {
    func mySqrt(_ x: Int) -> Int {
        //牛顿迭代法
        //判断x是否为0
        if x == 0 {return 0}     
        var low:Double = 1.0
        var high:Double = Double(x)
        var mid:Double = high/2
        var y:Double = (mid>1) ? mid : 1 
        while(true)
        {
            var dy:Double = (y * y + high) / Double(y)/2
            //处理int类型的取绝对值abs(int i)
            //处理double类型的取绝对值fabs(double i)
            //处理float类型的取绝对值fabsf(float i)
            if fabs(y - dy)<=0.01
            {
                return Int(dy)
            }
            else
            {
                y = dy
            }
        }
    }
}

---

12ms

class Solution {
  func mySqrt(_ x: Int) -> Int {
    return Int(Double(x).squareRoot())
  }
}

---

16ms

class Solution {
    func mySqrt(_ x: Int) -> Int {
        
        if x <= 1 {
            return x
        }
        
        var leftIndex = 1
        var rightIndex = x
        
        while leftIndex <= rightIndex {
            let middle = (leftIndex + rightIndex) / 2
            if middle * middle == x {
                return middle
            } else if middle * middle < x {
                leftIndex = middle + 1
            } else if middle * middle > x {
                rightIndex = middle - 1
            }
        }
        
        return rightIndex
    }
}

---

28ms

class Solution {
    func mySqrt(_ x: Int) -> Int {
        guard x > 1 else { return x }
        var left = 0
        var right = x
        while left <= right {
            let mid = (left + right) / 2
            if mid*mid > x {
                right = mid - 1
            }
            else {
                left = mid + 1
            } 
        }
        return right
    }
}