Implement 

int sqrt(int x)

Compute and return the square root of x.

求一个数的平方根。

解法：二分法，迭代循环在x范围内找中间值mid，然后判断mid * mid和x，如果mid  > x/mid（不要写成middle*middle==x，会溢出），说明这个数大了，就保留左边，right = mid -1。否则保留右边, left = mid + 1。直到left > right结束循环，返回left - 1。因为当x>2时，x/2的平方一定大于x，不可能是平方根，右指针可以从x/2开始。

Java:

public class Solution {
public int sqrt(int x) {
if(x<=1) {
return x;
}

int begin = 1;
int end = x;
int middle = 0;
while(begin<=end) {
mid = begin + (end - begin)/2;
if(middle == x/mid) {
return mid;
} else {
if (middle < x/mid) {
begin = mid + 1;
} else {
end = mid - 1;
}
}

}
return end;
}
} 　

Python:

class Solution(object):
def mySqrt(self, x):
"""
:type x: int
:rtype: int
"""
if x < 2:
return x

left, right = 1, x // 2
while left <= right:
mid = left + (right - left) // 2
if mid > x / mid:
right = mid - 1
else:
left = mid + 1

return left - 1

C++:

class Solution {
public:
int mySqrt(int x) {
if (x <= 1) return x;
int left = 0, right = x;
while (left < right) {
int mid = left + (right - left) / 2;
if (x / mid >= mid) left = mid + 1;
else right = mid;
}
return right - 1;
}
};

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