LeetCode 50. Pow(x, n)

1.タイトルの説明
Implement pow(x, n).
2.問題の解き方
主にいくつかの状況を考慮する必要があり、底数は-1、1、指数は0、正数、負数である.正常な指数が正の場合、このような数学のテクニックを使うことを考えることができます.
res(x,n)={res(x2,n/2)∗xres(x2,n/2)ifn%2==1else
3. code

class Solution {
public:
    double myPow(double x, int n) {
        if (n == 0)
            return 1.0;

        if (fabs(x - 1) < 1e-6)
            return 1.0;

        if (fabs(x + 1) < 1e-6)
            return n % 2 ? -1.0 : 1.0;

        if (n < 0){
            long long tmp = n;
            tmp = -tmp;
            return _myPow(1 / x, tmp);
        }

        return _myPow(x, n);
    }

    double _myPow(double x, long long n) {
        if (n == 1)
            return x;

        if (n % 2){
            return x * _myPow(x * x, n >> 1);
        }
        return _myPow(x * x, n >> 1);
    }
};

4.大神解法
4.1 demo1
彼はここで指数が0,正数,負数の場合を特殊に処理した.

public class Solution {
    public double pow(double x, int n) {
        if(n == 0)
            return 1;
        if(n<0){
            n = -n;
            x = 1/x;
        }
        return (n%2 == 0) ? pow(x*x, n/2) : x*pow(x*x, n/2);
    }
}

4.2 demo2

After reading some good sharing solutions, I'd like to show them together. You can see different ideas in the code.

1. nest myPow

double myPow(double x, int n) {
    if(n<0) return 1/x * myPow(1/x, -(n+1));
    if(n==0) return 1;
    if(n==2) return x*x;
    if(n%2==0) return myPow( myPow(x, n/2), 2);
    else return x*myPow( myPow(x, n/2), 2);
}
2. double myPow

double myPow(double x, int n) { 
    if(n==0) return 1;
    double t = myPow(x,n/2);
    if(n%2) return n<0 ? 1/x*t*t : x*t*t;
    else return t*t;
}
3. double x

double myPow(double x, int n) { 
    if(n==0) return 1;
    if(n<0){
        n = -n;
        x = 1/x;
    }
    return n%2==0 ? myPow(x*x, n/2) : x*myPow(x*x, n/2);
}
4. iterative one

double myPow(double x, int n) { 
    if(n==0) return 1;
    if(n<0) {
        n = -n;
        x = 1/x;
    }
    double ans = 1;
    while(n>0){
        if(n&1) ans *= x;
        x *= x;
        n >>= 1;
    }
    return ans;
}
5. bit operation

class Solution {
public:
    double pow(double x, int n) {
        if(n<0){
            x = 1.0/x;
            n = -n;
        }
        int unsigned m = n;
        double tbl[32] = {0};
        double result = 1;
        tbl[0] = x;
        for(int i=1;i<32;i++){
            tbl[i] = tbl[i-1]*tbl[i-1];
        }
        for(int i=0;i<32;i++){
            if( m & (0x1<<i) )
            result *= tbl[i];
        }
        return result;
    }
};

If you have other ideas, please leave it below. Thanks.