Pow(x, n) LeetCode solution in C++

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class Solution {
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 public:
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  double myPow(double x, int n) {
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    if (n == 0) return 1; // anything to the power 0 is 1
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    if (x == 1 || n == 1) return x; // 1 to the power anything = 1 or x to the power 1 = x
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    double ans = 1;
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    long long int a = abs(n); // since int range is from -2147483648 to 2147483647, so it
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                              // can't store absolute value of -2147483648
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    if (n < 0) { // as 2^(-2) = 1/2^2
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      if (a % 2 == 0)
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        ans = 1 / myPow(x * x, a / 2);
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      else
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        ans = 1 / (x * myPow(x, a - 1));
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    } else {
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      if (a % 2 == 0)
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        ans = myPow(x * x, a / 2);
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      else
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        ans = x * myPow(x, a - 1);
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    }
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    return ans;
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  }
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};