Sqrt (int x)

Question: Implement int sqrt (int x ). compute and return the square root of x. an1_1: bipartite [cpp] class Solution {public: int sqrt (int x) {if (x <0) return-1; // assert (x> = 0 ); long x2 = (long) x; long left = 0; long right = x2; long mid = 0; while (left <= right) {mid = left + (right-left)/2; if (mid * mid = x2 | (mid * mid <x2 & (mid + 1) * (mid + 1)> x2) {return (int) mid;} else if (mid * mid <x2) {left = mid + 1 ;} else {right = mid-1 ;}}}; Note: 1) non-negative number judgment, negative number does not have an open square root 2) value range, mid = left + (right-left) /2; may exceed the maximum int value range. Therefore, you must set the mid type to long (C ++ without ulong) an1_2: Newton iteration method [cpp] class Solution {public: int sqrt (int x) {if (x <0) return-1; // assert (x> = 0); double n = x; while (abs (n * n-x)> 0.0001) {n = (n + x/n)/2;} return (int) n ;}}; note: to solve the square root problem of a, we can convert it to x ^ 2-a = 0 to evaluate the x value, and then abc (x ^ 2-a) <0.0001 (0.0001 is near precision) f (x) = x ^ 2-a, f (x) is the accuracy value range (infinitely close to 0). Evaluate f (x) function: transformation formula: evaluate the f (x) = x ^ 2-a formula and import it to: Xn + 1 = Xn-(Xn ^ 2-a)/(2Xn) = Xn-(Xn-a/Xn)/2 = (Xn + a/Xn)/2 where Xn + 1 is infinitely close to Xn, I .e: xn = (Xn + a/Xn)/2 anw.3: Martian algorithm [cpp] # include <stdio. h> int InvSqrt (int x) {float x2 = (float) x; float xhalf = x2/2; int I = * (int *) & x2; // get bits for floating VALUE I = 0x5f375a86-(I> 1); // gives initial guess y0 x2 = * (float *) & I; // convert bits BACK to float x2 = x2 * (1.5f-xhalf * x2 * x2); // Newton step, repeating increases accuracy x2 = x2 * (1.5f-xhalf * x2 * x2); // Newton step, repeating increases accuracy x2 = x2 * (1.5f-xhalf * x2 * x2 ); // Newton step, repeating increases accuracy printf ("\ n \ n1/x = % d \ n", (int) (1/x2); return (int) (1/x2);} int main () {// InvSqrt (65535); InvSqrt (10); InvSqrt (2147395599); InvSqrt (1297532724); return 0;} description: this method is highly efficient. I use the int (parameter) written by someone else's float)