15.Pow(x, n)
- Pow(x, n)
Total Accepted: 88351 Total Submissions: 317095 Difficulty: Medium
Implement pow(x, n).
思路:a.循环迭代求解(O(n)n次乘法运算)
仔细想想,做了很多重复工作
b.如何利用已经结论呢
28=2∗2∗2....∗2 28=(24)∗ans(ans表示前面已经计算好的24)
思路来了,二分,分奇偶讨论,复杂度为O(log(n))
Input:2.00000 -2147483648
Output:inf
Expected:0.00000
#define IMIN numeric_limits<int>::min()
class Solution {
public:
double myPow(double x, int n) {
int is_min=0;
if(n<0){
x=1/x;
if(n==IMIN){n=abs(n+1);is_min=1;}//处理负数绝对值越界,并标记
else n=abs(n);
}
if(n==0)return 1;
int sign=-1;
if(n%2==0)sign=0;//判断奇偶
else sign=1;
double left = myPow(x,n/2);//二分乘法
double res=0;
if(sign==0)res= left*left;
else res= left*left*x;
if(is_min)res = res*x;//为负数边界转为正数,少乘了一个
return res;
}
};