Leetcode: Sqrt (x)

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I'm so happy. Let's just take two points. However, it should be noted that the use of long is cross-border TM, and the use of unsigned long. Finally, check the returned value.

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;    }

For ease of understanding, let's take this question as an example:

Iteration formula_{The program is easy to write. For more information about the Newton Iteration Method, see.}

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;}