Next time you see a numeric problem has a too straightforward solution, think about optimized one.
Like this one: recursion\iteration is tooo slow. So Dichotomy, like sqrt(). Take care of minux n case.
class Solution { public: double pow(double x, int n) { if(n == 0) return 1; bool bNeg = false; if(n < 0) { bNeg = true; n *= -1; } double r = pow(x, n/2); if(n % 2 == 0) r *= r; else r *=r * x; if(!bNeg) return r; else return 1.0/r; } };
LeetCode "Pow(x,n)",布布扣,bubuko.com
原文：http://www.cnblogs.com/tonix/p/3884032.html