Jan 19-sqrt (x); Math; Binary Search;

We should keep in mind so never let the integer value overranged or overstacked.

Here are the trick, instead of comparing the square of a interger with the target value, we can compare the integer value I With the quotient x/i;

if (i > x/i) high = i-1; if (I < x/i) low = i +1, the condition of the and is loop are low < high. Thus when the loop is ends, low = high. There is and results, one is the low point to the integer which was the sqrt (x), the other is low = sqrt (x) +1. We need to check the "cases" in the end.

Code:

public class Solution {public    int mysqrt (int x) {        if (x<0) return-1;        if (x = = 0) return 0;        if (x = = 1) return 1;        int low = 1, high = x;        while (Low < high) {            int mid = low+ (high-low)/2;            int quotient = X/mid;            if (quotient = = mid) return mid;            else if (Mid > Quotient) {high                = mid-1;            }            else{Low                = mid + 1;            }        }        int quotient = X/low;        if (Low > Quotient) return low-1;        return low;    }}

　　

Newton ' s method to get square root:

Code:

 Public int mysqrt (int  x) {    ifreturn 0;     long i = x;      while (i > x/ i)           = (i + x/i)/2;                 return (int) i;}

Jan 19-sqrt (x); Math; Binary Search;