力扣69、x的平方根

1、换函数exp()(0ms,100%;6.2MB,5%)

1  int mySqrt(int x) {
2         if(x==0)
3         return x;
4        int num=exp(0.5*log(x));
5        if(long (num+1)*(num+1)<=x)
6        return num+1;
7        else
8        return num;
9     }

2、二分查找(4ms,59%;5.8MB,64%)

 1 int mySqrt(int x) {
 2         if(x<=1)
 3         return x;
 4         long left=1,right=x;
 5         long mid=(left+right)/2;
 6         while(left<=right){
 7             //因为很多数的算数平方根是夹在两个数中间的
 8             if(((mid*mid)<=x)&&((mid+1)*(mid+1)>x))
 9             return mid;
10             //mid偏小，整体往右移动
11             else if((mid*mid)<x){
12                 left=mid+1;
13                 mid=(left+right)/2;
14             }
15             //mid偏大，整体往右移动
16             else {
17                 right=mid-1;
18                 mid=(left+right)/2;
19             }
20             
21         }
22         return 0;
23     }

3、牛顿迭代（4ms,59%;5.8MB,60%）:没看懂

 1 int mySqrt(int x) {
 2        if (x == 0) {
 3             return 0;
 4         }
 5 
 6         double C = x, x0 = x;
 7         while (true) {
 8             double xi = 0.5 * (x0 + C / x0);
 9             if (fabs(x0 - xi) < 1e-7) {
10                 break;
11             }
12             x0 = xi;
13         }
14         return int(x0);
15     }

标签：xi,return,int,mid,力扣,num,69,平方根,x0
来源： https://www.cnblogs.com/Wow-A/p/15610689.html