问题：
实现次方运算
Implement pow(x, n).
解法：
Consider the binary representation of n. For example, if it is "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. To speed up, we loop through each bit, if the i-th bit is 1, 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.
n还大于0的时候，每次循环x都在平方。遇到位为1的时候，把x乘进去。
Java代码：
public static double myPow(double x, int n) { if (n == 0) { return 1; } if (n < 0) { if (n == Integer.MIN_VALUE) { return 1.0 / (myPow(x, Integer.MAX_VALUE)*x); } else { return 1.0 / (myPow(x,-n)); } } double res = 1.0; for (;n > 0;x *= x,n>>=1) { if ((n & 1) > 0) { res *= x; } } return res; }
public static double myPow(double x, int n) { if (n == 0) { return 1; } if (n < 0) { if (n == Integer.MIN_VALUE) { return 1.0 / (myPow(x, Integer.MAX_VALUE)*x); } else { return 1.0 / (myPow(x,-n)); } } double res = 1.0; while (n > 0) { if ((n & 1) > 0) { res *= x; } x *= x; n>>=1; } return res; }
