Implement int sqrt(int x).
Compute and return the square root of x.
参考：http://standalone.iteye.com/blog/1847368
参考的是一个用二分查找实现的，这道题还可以用什么牛顿法之类的
如果middle * middle > x在左半部分查找，如果middle * middle < x在有半部分查找。初始状态low = 0, high = x
1 public class Solution { 2 public int sqrt(int x) { 3 double low = 0; 4 double precision = 0.000000001; 5 double high = x; 6 while((high - low ) >precision){ 7 double middle = (low + high) / 2; 8 if(x > middle * middle){ 9 low = middle; 10 } 11 if(x < middle * middle){ 12 high = middle; 13 } 14 if(x == middle * middle){ 15 return (int)Math.floor(middle); 16 } 17 }//while 18 return (int)Math.floor(high); 19 } 20 }
原文：http://www.cnblogs.com/luckygxf/p/4219054.html