leetcode50 Pow(x, n)

50. Pow(x, n)

Total Accepted:
83619 Total Submissions:
299730 Difficulty: Medium
Implement pow(x, n).

二分法求快速幂，时间复杂度为O（logn）。

测试数据有各种负数，真是醉了

最初的AC代码：

class Solution {
public:
    double myPow(double x, int n) {
        
        if (0 == n || 1 == x) return 1;
        if (-1 == x) {
            if (n & 1)
                return -1;
            return 1;
        }
        if (-2147483648 == n) return 0;
        
        double ret = 1;
        int k = abs(n);
        while (k) {
            if (k & 1) {
                ret *= x;
            }
            x *= x;
            k >>= 1;
        }
        if (n > 0) return ret;
        return 1 / ret;
    }
};

改进后的AC代码：

class Solution {
public:
double myPow(double x, int n) {

if (0 == n) return 1;
if (INT_MIN == n) n += 2; //n绝对值比较大，加2避免溢出，也无伤大雅

double ret = 1;
int k = abs(n);
while (k) {
if (k & 1) {
ret *= x;
}
x *= x;
k >>= 1;
}
if (n > 0) return ret;
return 1 / ret;
}
};