leetcode-50-Pow(x, n)

最新推荐文章于 2025-07-28 18:52:14 发布

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本文深入探讨了使用递归算法高效计算幂运算的方法。首先判断指数是否为负数，若为负则转换为正数并计算其倒数。接着检查指数是否为偶数，若是，则计算底数平方的幂；若否，则计算底数平方的幂再乘以底数。此方法基于幂的性质，即较大基数的小幂等于较小基数的大幂。

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Use recursive to deal with this. First if n < 0 then calculate its positive one then 1 / its reuslt, then, check if current n is even, it it is, calculate its pow(x*x), if not, calculate its pow(x * x) *x.

The idea behind it is power of something is equal its bigger base power of lower power, i.e.
x^n = (x^ 2)^(n/2)

Error:
1. Do not take care of the number overflow, always remember, if it has something about the type, abs(INT_MIN) = abs(INT_MAX + 1), so sometimes we will meet the error.