【题目】
多次方计算，主要是正负两种情况
Implement pow(x, n), which calculates x raised to the power n (i.e., xn).
Example 1:
Input: x = 2.00000, n = 10 Output: 1024.00000
Example 2:
Input: x = 2.10000, n = 3 Output: 9.26100
Example 3:
Input: x = 2.00000, n = -2 Output: 0.25000 Explanation: 2-2 = 1/22 = 1/4 = 0.25
【思路】
- 分类讨论正负两种情况即可
- 将负数也全部转换为正数计算，即x^-1=1/x
- 避免负数时 2.00000 -2147483648溢出
【代码】
class Solution { public double myPow(double x, int n) { if(n == 0) return 1; if(n<0){ return 1/x * myPow(1/x, -(n + 1)); //1/x 因为x^-1=1/x //-(n+1)变成正数处理，把负数给x了 } return (n%2 == 0) ? myPow(x*x, n/2) : x*myPow(x*x, n/2); } }