Leetcode: Sqrt(x)

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就高興了，直接二分吧。但是要注意，即使用long long都TM越界，還要用unsigned long long。最後返回值還要再檢查一下。

int sqrt(int x) {        // Start typing your C/C++ solution below        // DO NOT write int main() function        unsigned long long begin = 0;        unsigned long long end = (x+1)/2;        unsigned long long mid;        unsigned long long tmp;        while(begin < end)        {            mid = begin + (end-begin)/2;            tmp = mid*mid;            if(tmp==x)return mid;            else if(tmp<x) begin = mid+1;            else end = mid-1;        }        tmp = end*end;        if(tmp > x)            return end-1;        else            return end;    }

   為了方便理解，就先以本題為例：

迭代公式_{，程式就好寫了。關於牛頓迭代法，可以參考以及。}

int sqrt(int x) {// Start typing your C/C++ solution below        // DO NOT write int main() function        if (x ==0)            return 0;        double pre;        double cur = 1;        do        {            pre = cur;            cur = x / (2 * pre) + pre / 2.0;        } while (abs(cur - pre) > 0.00001);        return int(cur);    }

float InvSqrt(float x){    float xhalf = 0.5f*x;    int i = *(int*)&x; // get bits for floating VALUE    i = 0x5f375a86- (i>>1); // gives initial guess y0    x = *(float*)&i; // convert bits BACK to float    x = x*(1.5f-xhalf*x*x); // Newton step, repeating increases accuracy    return x;}