Leetcode | | 50. Pow (x, N)

Problem:

Implement Pow (x, n).

Hide TagsMath Binary SearchTest instructions: To find the n power of X

Thinking:

(1) The simplest thought is the intuitive mathematical power function to find the method, the test passed. Algorithm time complexity of O (n)

(2) Follow the label prompts, using the binary search method. Pow (x,n) = POW (X,N-N/2) *pow (X,N/2), each scaling the size of n, attention to the parity of N is discussed, the algorithm time complexity is log (n)

(3) In addition to the above method, a very ingenious and fast method is also mentioned here, which is described as follows:

Consider the binary representation of N. For example, if it was "10001011", then X^n = x^ (1+2+8+128) = x^1 * x^2 * x^8 * x^128. Thus, we don ' t want to loop n times to calculate x^n. We loop through each bit, and if the i-th bit is 1 and then we add x^ (1 << i) to the result. Since (1 << i) is a power of 2, x^ (1<< (i+1)) = Square (x^ (1<<i)). The loop executes for a maximum of log (n) times.

The method calculates the power of x by scanning 1 of different positions in the binary representation of N, with the worst being O (n), but with a good average complexity

Code

(1) Recursive method: Accepted

Class Solution {public:    double pow (double x, int n) {        double ret=1.0;        if (x==1.0)            return 1.0;        if (x==-1.0)        {            if (n%2==0)                return 1.0;            else                return-1.0;        }        if (n<0)            return 1.0/pow (x,-n);        while (n)        {            if (ret==0)//Prevent run timeout                return 0;            ret*=x;            n--;        }        return ret;    }};

(2) Dichotomy: Accepted

Class Solution {public:    double pow (double x, int n) {        //double ret=1.0;        if (x==1.0)            return 1.0;        if (x==-1.0)        {            if (n%2==0)                return 1.0;            else                return-1.0;        }        if (n==0)            return 1.0;        if (n<0)            return 1.0/pow (x,-n);        Double Half=pow (x,n>>1);        if (n%2==0)            return half*half;          else            return x*half*half;}    };

(3)

In order to properly calculate the n power of X, there are a few things to consider:

1) x is 0 o'clock, the power of 0 is 1, and negative power is meaningless; judging whether x equals 0 cannot be used directly with "= =".

2) when n is Int_min,-n is not Int_max, this requires extra care.

3) Try to use shift operations instead of division operations to speed up the execution of the algorithm.

Class Solution {public:    double pow (double x, int n) {        //Start Typing your C + + solution below        //Do not writ e int main () function        if (n<0) {if (n==int_min) return 1.0/(POW (X,int_max) *x); Elsereturn 1.0/pow (x,-n);}        if (n==0)            return 1.0;double ans = 1.0; for (;n>0; x *= x, n>>=1) {if (n&1>0) ans *= x;} return ans;    };

Leetcode | | 50. Pow (x, N)