Implement int sqrt(int x).
Compute and return the square root of x, where x is guaranteed to be a non-negative integer.
Since the return type is an integer, the decimal digits are truncated and only the integer part of the result is returned.
Example 1:
Input: 4 Output: 2
Example 2:
Input: 8 Output: 2 Explanation: The square root of 8 is 2.82842..., and since the decimal part is truncated, 2 is returned.
代码：
用反向的思想，i从1开始试，判断啥时候i*i能大于所给的数。
这里要注意最大值问题，计算机能表示的数最大是2147483647，而46340*46340 = 2 147 395 600， 也就是说46341 的平方会超出计算机所能表示的数，所以我们设置不管多大的数，算到46340就不往下算了。
class Solution {
public:
int mySqrt(int x) {
int i = 1;
while(i*i<=x&&i!=46340)i++;
if(i*i>x)
return i-1;
else
return i;
}
};