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SqrtOfNumber.java
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SqrtOfNumber.java
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package Leetcode;
/**
* @author kalpak
*
* 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.
*
*
* 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 <= 2^31 - 1
*/
public class SqrtOfNumber {
public static int mySqrt(int x) {
if(x < 2) {
return x;
}
int left = 1;
int right = x;
int result = 0;
while(left <= right) {
int mid = left + (right - left) / 2;
if(mid > x/mid) {
// this means mid^2 > x, so reduce the search range.
// avoid squaring mid to avoid overflow.
right = mid - 1;
}
else {
left = mid + 1;
result = mid; // possible candidate. But can be better.
}
}
return result;
}
public static void main(String[] args) {
System.out.println(mySqrt(26));
}
}