题目描述

• 很棒的题目！便于分治思想的理解！
  

思路 && 代码

• 分情况是大头～
• 递归结束的情况：
  1. $x^0$ = 1
  2. $x^1$ = x
  3. x${}^{-}1$ = 1 / x
• 偶数情况：直接二分递归
• 奇数情况：先向下取整二分递归，再补乘一个n即可

class Solution {// 正负数都满足情况public double myPow(double x, int n) {// 三个递归结束条件if(n == 0) { return 1; }if(n == 1) { return x; }if(n == -1) { return 1 / x; }// 递归代码double half = myPow(x, n / 2);// half 向下取整，加一个奇数情况判断double mod = myPow(x, n % 2);// x^(n/2) * x^(n/2) * x^(0 or 1) = x^nreturn half * half * mod;}
}

二刷

class Solution {public double myPow(double x, int n) {if(n == 0) {return 1;}if(n == 1) {return x;}if(n == -1) {return 1 / x;}double half = myPow(x, n / 2);return half * half * myPow(x, n % 2);}
}