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。
版本：二分法用除法（20ms）
1 class Solution {
2 func mySqrt(_ x: Int) -> Int {
3 //二分法
4 var low:Int = 1
5 var high:Int = x
6 while(low <= high)
7 {
8 var mid = (low+high)/2
9 //乘法会导致溢出，应写成除法
10 if mid == x/mid
11 {
12 return mid
13 }
14 else
15 {
16 if mid < x/mid
17 {
18 low = mid + 1
19 }
20 else
21 {
22 high = mid - 1
23 }
24 }
25 }
26 return low-1
27 }
28 }
版本：二分法用乘法（24ms），可能导致溢出
1 class Solution {
2 func mySqrt(_ x: Int) -> Int {
3 if x < 2 { return x }
4
5 var low = 1
6 var high = x / 2
7
8 while (low <= high) {
9 let mid = low + (high - low) / 2
10 if mid * mid <= x && (mid+1) * (mid+1) > x {
11 return mid
12 } else if mid * mid <= x && (mid+1) * (mid+1) <= x {
13 low = mid + 1
14 } else {
15 high = mid - 1
16 }
17 }
18
19 return -1
20 }
21 }
牛顿迭代法：20ms
1 class Solution {
2 func mySqrt(_ x: Int) -> Int {
3 //牛顿迭代法
4 //判断x是否为0
5 if x == 0 {return 0}
6 var low:Double = 1.0
7 var high:Double = Double(x)
8 var mid:Double = high/2
9 var y:Double = (mid>1) ? mid : 1
10 while(true)
11 {
12 var dy:Double = (y * y + high) / Double(y)/2
13 // 处理int类型的取绝对值abs(int i)
14 //处理double类型的取绝对值fabs(double i)
15 //处理float类型的取绝对值fabsf(float i)
16 if fabs(y - dy)<=0.01
17 {
18 return Int(dy)
19 }
20 else
21 {
22 y = dy
23 }
24 }
25 }
26 }