更新时间:2022-04-18 18:28:51
Implement pow(x, n).
其实这个题递归和非递归应该都不难实现,就是边界值的问题。
public double myPow(double x, int n) {
if (n == 0)
return 1.0;
double res = 1.0;
if (n < 0) {
if (x >= 1.0 / Double.MAX_VALUE || x <= 1.0 / -Double.MAX_VALUE)
x = 1.0 / x;
else
return Double.MAX_VALUE;
if (n == Integer.MIN_VALUE) {
res *= x;
n++;
}
}
n = Math.abs(n);
boolean isNeg = false;
if (n % 2 == 1 && x < 0) {
isNeg = true;
}
x = Math.abs(x);
while (n > 0) {
if ((n & 1) == 1) {
if (res > Double.MAX_VALUE / x)
return Double.MAX_VALUE;
res *= x;
}
x *= x;
n = n >> 1;
}
return isNeg ? -res : res;
}