Leetcode 69. Sqrt(x)

原题：

Implement

int sqrt(int x)

.

Compute and return the square root of x.

x is guaranteed to be a non-negative integer.

Example 1:

Input: 4
Output: 2

Example 2:

Input: 8
Output: 2
Explanation: The square root of 8 is 2.82842..., and since we want to return an integer, the decimal part will be truncated.

解决方法：

从该数本身开始，找到第一个平方不小于目标的数即是所求解。

代码：

int mySqrt(int x) {
if(x < 0)
return -1;
if (x <=1)
return x;
long i = x;
while(i*i > x){
i = ((i + x/i) >> 1);
}
return i;
}