[Leetcode] SQRT (X)

ImplementInt SQRT (int x).

Compute and return the square rootX.

Https://oj.leetcode.com/problems/sqrtx/

 

Idea 1: the square root of a number must be 0 ~ X/2 + 1, so binary search in this range.

Note:

• Long is used in the intermediate result; otherwise, mid * mid overflows.
• Returns the floor value of the square root if no result is found.

Idea 2: Newton Iteration Method.

 

 Public   Class  Solution {  Public   Int SQRT ( Int X ){  If (X <0 )  Return -1 ;  Long Right = X/2 + 1 ;  Long Left = 0 ;  Long  Mid;  While (Left <= Right) {mid = (Left + right)/2;  If (X <mid * Mid) Right = Mid-1 ;  Else   If (X> mid * Mid) left = Mid + 1 ;  Else                  Return ( Int  ) Mid ;}  Return ( Int ) Right ;}  Public   Static   Void  Main (string [] ARGs) {system. Out. println (  New Solution (). SQRT (0 ); System. Out. println (  New Solution (). SQRT (1 ); System. Out. println (  New Solution (). SQRT (2 ); System. Out. println (  New Solution (). SQRT (3); System. Out. println (  New Solution (). SQRT (4 ); System. Out. println (  New Solution (). SQRT (2147483647 ));}} 

View code

 

Refer:

Http://blog.csdn.net/linhuanmars/article/details/20089131

Http://www.cnblogs.com/AnnieKim/archive/2013/04/18/3028607.html