In this **Leetcode Sqrt(x) problem solution** we have Given a non-negative integer x, compute and return the square root of x. Since the return type is an integer, the decimal digits are truncated, and only the integer part of the result is returned.

## Problem solution in Python.

class Solution: def mySqrt(self, x: int) -> int: return floor(sqrt(x))

## Problem solution in Java.

public class Solution { public int mySqrt(int x) { int left=1,right =x, result=0; while(left<right){ int mid = left + (right - left)/2; if(mid<x/mid){ left = mid+1; }else{ right = mid; } } return (left==x/left)?left:left-1; } }

## Problem solution in C++.

class Solution { public: int mySqrt(int x) { int y = 1; int n = x; while (n > y) { n = y + (n - y) / 2; y = x / n; } return n; } };

## Problem solution in C.

long int mySqrt(int x){ long int i=1,rt; if(x==0) return 0; for(i;i*i<=x;i++) { rt=i; } return rt; }

## 0 Comments