LeetCode 69. Sqrt(x)

class Solution {public:    int mySqrt(int x) {        double cur = x,pre =0;        while(abs(pre-cur)>0.000001){            pre = cur;            cur = pre/2 + (x/(2*pre));        }        return int(cur);    }};

以下代码cur = pre/2 + x/(2*pre)是化简计算的结果。。这里的f(x) = x^2-n

f'(x) = 2*x = 2*pre

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