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.
Tags: Binary Search, Math
题意是求平方根，参考 牛顿迭代法求平方根，然后再参考维基百科的 Integer square root 即可。
class Solution {
public int mySqrt(int x) {
long n = x;
while (n * n > x) {
n = (n + x / n) >> 1;
}
return (int) n;
}
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