[leetcode]Pow(x, n)

借鉴了之前加法的超时经验，就开始采用倍增法。但还是吃了负数和整数边界值的亏。最后干脆使用long得了。参考答案的递归果然更简洁易懂，而且不用考虑整数边界值的情况，精彩。

public class Solution {
    public double pow(double x, int n) {
        // Start typing your Java solution below
        // DO NOT write main() function
        
        double ans = 1;
    	double tmp = x;
        boolean neg = false;
        long m = n;
        if (m < 0) {
            neg = true;
            m = -m;
        }
    	long bound = m;
    	
    	while (bound != 0) {
    		long i = 1;
	    	for (; i*2 <= bound; i*=2) {
	    		tmp = tmp * tmp;
	    	}
	    	ans *= tmp;
	    	tmp = x;
	    	bound = bound - i;
    	}
        
        if (neg) return 1.0 / ans;
    	
    	return ans;
    }
}

参考答案：http://discuss.leetcode.com/questions/228/powx-n

double pow(double x, int n) {
    if (n == 0) return 1.0;
    // Compute x^{n/2} and store the result into a temporary
    // variable to avoid unnecessary computing
    double half = pow(x, n / 2);
    if (n % 2 == 0)
        return half * half;
    else if (n > 0)
        return half * half * x;
    else
        return half * half / x;
}

第二刷：要注意n是负数的情况，还有负到头~

class Solution {
public:
    double pow(double x, int n) {
        if (x == 0 || x == 1) return x;
        if (n < 0 && n != INT_MIN) return 1.0 / pow(x, -n);
        if (n == 0) return 1.0;
        double p = pow(x, n / 2);
        if (n % 2 == 0) {
            return p * p;
        } else {
            return p * p * x;
        }
    }
};