Pow (x, N)--Leetcode

Implement POW (double x, int n). –leetcode
Custom implementation POW function.
Note the point:
(1) The value of double type, cannot use = = to judge whether equal, should use a certain error value to judge;
(2) Handling of special cases:
1) x is 0 and n is 0
2) x is not 0, N is 0
3) x is not 0, N is a negative case. (You can convert negative numbers to positive numbers and then the inverse of the results.) Note: The minimum value of a negative number, converted to positive, will cross the border. The solution is to add 1 to the negative, then to positive, then multiply 1/x to ensure the result.

If the process of iterative iteration is solved directly, the time complexity is too high and it is difficult to compile. Observation can know:
When n > 0 o'clock:

According to the iteration formula-there is code:

 Public:Double Power(Double Base,intExponent) {The //base index is 0, equal (base,0) is to determine whether the bases and 0 are equal (special case processing, here when the radix index is 0, the value is 1)        if(Equal (Base,0) && exponent = =0)return 1.0;//index is 0        if(Exponent = =0)return 1.0;//exponent is plural       if(Exponent <0){//exponent is negative, exponent plus 1, then convert to positive           return 1/Base* Powerwithexponent (1/Base,-(exponent+1)); }//exponent is positive       returnPowerwithexponent (Base, exponent); }//recursive solution to think:    //a^x = (1) a^ (X/2) * a^ (X/2) x is even    //(2) a^ ((x-1)/2) * a^ ((x-1)/2) * A x is odd    DoublePowerwithexponent (Double Base,intExponent) {//index is 0, return 1        if(Exponent = =0)return 1;//index is 1, back to base        if(Exponent = =1)return Base;Doubleresult = Powerwithexponent (Base, exponent >>1);//exponent >> 1 bit shift operation, equivalent to dividing by 2, but better performanceResult *= result;if(Exponent &0x1){//exponent & 0x1 is equivalent to the% operation, better performance. Exponent & 0x1 = = 1 means oddResult *=Base; }returnResult }//Determine if two double is equal and cannot be solved with = =    BOOLEqual (DoubleADoubleb) {if(A-B >-0.0000001) && (A-B) <0.0000001)return true;Else return false; }    };

Pow (x, N)--Leetcode