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;
}
}