LeetCode_50_Pow(x, n)

题目：Implement pow(x, n).

解法1：常规递归，时间复杂度o(n)

解法2：分治法改进递归，注意 n == INT_MIN 情况需单独考虑。尽量直接return而不要赋值给变量返回，容易出问题

double myPow(double x, int n) {
if (n == 0)
return 1.0;
else if (n > 0)
{
double half = (double)myPow(x, n / 2);
if (n % 2 == 0)
return half*half;
else
return half*half*x;
}
else
{
if (n == INT_MIN) {
return 1.0 / (pow(x, INT_MAX)*x);
//return 1.0 / (half*half);
}

n = -n;

double half = (double)myPow(x, n / 2);

if (n % 2 == 0)
return 1.0 / (half*half);
else
return 1.0 / (half*half*x);
}
}