LeetCode: Pow(x, n)

早早投降了,看了这个强大的代码

 1 class Solution {
 2 public:
 3     double pow(double x, int n) {
 4         // Start typing your C/C++ solution below
 5         // DO NOT write int main() function
 6         bool negtive = n < 0;
 7         n = abs(n);
 8         double ret = 1.0;
 9         while (n > 0) {
10             ret = (n & 1) == 1? ret*x : ret;
11             x *= x;
12             n >>= 1;
13         }
14         return negtive? 1/ret : ret;
15     }
16 };

 C# 注意下Int32.MinValue问题就行

 1 public class Solution {
 2     public double MyPow(double x, int n) {
 3         bool neg = n < 0;
 4         bool minValue = false;
 5         if (n == Int32.MinValue) {
 6             n = Int32.MaxValue;
 7             minValue = true;
 8         }
 9         else n = Math.Abs(n);
10         double ans = 1.0;
11         double x1 = x;
12         while (n > 0) {
13             ans = (n & 1) == 1? ans * x1 : ans;
14             x1 *= x1;
15             n >>= 1;
16         }
17         if (minValue) ans *= x;
18         return neg? 1 / ans : ans;
19     }
20 }
View Code
原文地址:https://www.cnblogs.com/yingzhongwen/p/3030893.html