int sqrt(int x).Compute and return the square root of x.
public class Solution { public int mySqrt(int x) { if(x == 0) return 0; int lowerBound = 1; int higherBound = 2; //search for a boundary while (true) { int square = lowerBound * lowerBound; if (square == x) return lowerBound; if (square < x) { lowerBound = higherBound; higherBound = higherBound * 2; } else { lowerBound = lowerBound / 2; higherBound = higherBound / 2; break; } } int left = lowerBound + 1; int right = higherBound - 1; while (left <= right) { int mid = (left + right) / 2; int square = mid * mid; if (square == x) return mid; if (square < x) left = mid + 1; else right = mid - 1; } return left - 1; } }
Another solution if start from both ends of Integer range.
public class Solution { public int mySqrt(int x) { if (x == 0) return 0; int left = 1, right = Integer.MAX_VALUE; while (left <= right) { int mid = left + (right - left)/2; int temp = x/mid;//use division because mid*mid may > Integer.max if(mid == temp) return mid; if (mid > temp) right = mid - 1; else left = mid + 1; } return left-1; } }
Given a positive integer num, write a function which returns True if num is a perfect square else False.
Note: Do not use any built-in library function such as sqrt.
Example 1:
Input: 16 Returns: True
Example 2:
Input: 14
Returns: False
public class Solution { public boolean isPerfectSquare(int num) { if (num == 0) return true; int left = 1, right = Integer.MAX_VALUE; while (left <= right) { int mid = left + (right - left)/2; int temp = num/mid;//use division because mid*mid may > Integer.max if(mid == temp) { if(mid*mid == num) return true; return false;//Here is the trap! e.g. num = 5, mid = 2 } if (mid > temp) right = mid - 1; else left = mid + 1; } return false; } }
69. Sqrt(x) && 367. Valid Perfect Square
原文：http://www.cnblogs.com/neweracoding/p/5686863.html