# Sqrt(x)

By [Vincentkovsky](https://paragraph.com/@wenzheng) · 2022-03-25

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**Binary search**

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.

Note: You are not allowed to use any built-in exponent function or operator, such as pow(x, 0.5) or x \*\* 0.5.

Example 1:

Input: x = 4 Output: 2 Example 2:

Input: x = 8 Output: 2 Explanation: The square root of 8 is 2.82842..., and since the decimal part is truncated, 2 is returned.

Constraints:

0 <= x <= 231 - 1

因为测试可能会出现input为2147395599这种情况，所以我们应当定义变量为long类型，避免overflow

    class Solution {
        public int mySqrt(int x) {
            if(x==0 || x==1){
                return x;
            }
            long left = 1;
            long right = x;
            
            while(left < right){
                long mid = left + (right - left) / 2;
                if(mid*mid == x){
                    return (int)mid;
                } else if(mid*mid > x){
                    right = mid;
                } else if(mid*mid < x){
                    left =mid+1;
                }
            }
            return (int)left-1;
        }
    }

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*Originally published on [Vincentkovsky](https://paragraph.com/@wenzheng/sqrt-x)*