# 69. Sqrt(x) (Easy)

https://leetcode.com/problems/sqrtx/

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.
```

## Solutions

``````class Solution {
public int mySqrt(int x) {
if (x == 1) {
return 1;
}

long least = 0;
long most = x;

while (least <= most) {
long mid = (most * 1l + least * 1l) / 2;
long squire = mid * mid;

if (squire == x) {
return (int) mid;
} else if (squire > x) {
most = (int) mid - 1;
} else {
// if mid is the largest integer of which the squire is the closest to but less than x
least = (int) (mid + 1);
}
}

// to reach here, least > most and most^2 < x < least^2
return (int) most;
}
}
``````