LeetCode

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.


使用暴力的方法 迭代，时间复杂度为O(n)
使用二分的方法，时间复杂度为O(logn)

 1 public int mySqrt(int x) {//二分 mytip
 2         if(0==x||1==x)
 3             return x;
 4         int left = 0;
 5         int right =x;
 6         while(left<=right){
 7             int mid = left+(right-left)/2;//可以使用(left+right)/2，但left或right过大时可能溢出
 8             int y = x/mid;//mid*mid越界
 9             if(mid<=y){
10                 int y1= x/(mid+1);
11                 if(y==mid||(mid+1)>y1){
12                     left =mid;
13                     break;
14                 }
15                 left=mid+1;
16             }
17             else{
18                 right=mid-1;
19             }
20         }
21         return left;
22     }

使用牛顿迭代法 https://blog.csdn.net/ccnt_2012/article/details/81837154

x_n+1 = x_n-f(x_n)/f'(x_n) = x_n-(x²-y₀)/(2*x_n)=(x_n+y₀/x_n)/2

 1 public int mySqrt(int x) {//牛顿迭代法 mytip
 2         if(0==x||1==x){
 3             return x;
 4         }
 5         long r=x;
 6         while(r>x/r){
 7             r = (r+x/r)/2;
 8         }
 9         return (int)r;
10     }

相关题

有效的完全平方数 LeetCode367 https://www.cnblogs.com/zhacai/p/10631678.html

扩展

根据精确度，返回double类型的结果

public double mySqrt1(double x,double pre) {
        double left = 0;
        double right =x;
        while((right-left)>pre){
            double mid = left+(right-left)/2;//可以使用(left+right)/2，但left或right过大时可能溢出
            double y = x/mid;//mid*mid越界
            if(y ==mid){
                left=mid;
                right = mid;
                break;
            }
            else if(y>mid){
                left=mid;
            }
            else{
                right=mid;
            }
        }
        //System.out.println(left);
        //System.out.println((left+right)/2);
        return left+(right-left)/2;
    }