leetcode -- Pow(x, n)

Implement pow(x, n).

直接计算会超时TLE

public double pow(double x, int n) {
        // Start typing your Java solution below
        // DO NOT write main() function
        if(n == 0)
            return 1;
        
        double result = 1;
        if(n > 0){
            while(n > 0){
                result *= x;
                n --;
            }
        } else {
            while(n < 0){
                result /= x;
                n ++;                
            }
        }
        return result;
    }

// TLE test case
0.00001, 2147483647

使用二分法来求解，O(lgn)

这里需要注意的是当n = -2147483648，0-n=-2147483648 即整数的整数范围最大为：-2147483648~2147483647

这里需要把int 提升为long

public class Solution {
    public double pow(double x, int n) {
        // Start typing your Java solution below
        // DO NOT write main() function
        long tmp = n;
        if(tmp >= 0){
            return power(x, tmp);
        } else {
            return 1.0 / power(x, -tmp);
        }
    }
    
    public double power(double x, long n){
        if(n == 0)
            return 1;
            
        if(n == 1)
            return x;
            
        double result = power(x, n >> 1);
        result *= result;
        if((n & 0x1) == 1){
            result = result * x;
        }
        
        return result;
    }
}