C++实现LeetCode(50.求x的n次方)

[LeetCode] 50. Pow(x, n) 求x的n次方

Implement pow(x, n), which calculates x raised to the power n(xn).

Example 1:

Input: 2.00000, 10
Output: 1024.00000

Example 2:

Input: 2.10000, 3
Output: 9.26100

Example 3:

Input: 2.00000, -2
Output: 0.25000
Explanation: 2-2 = 1/22 = 1/4 = 0.25

Note:

  • -100.0 < x < 100.0
  • n is a 32-bit signed integer, within the range [−231, 231 − 1]

这道题让我们求x的n次方，如果只是简单的用个 for 循环让x乘以自己n次的话，未免也把 LeetCode 上的题想的太简单了，一句话形容图样图森破啊。OJ 因超时无法通过，所以需要优化，使其在更有效的算出结果来们可以用递归来折半计算，每次把n缩小一半，这样n最终会缩小到0，任何数的0次方都为1，这时候再往回乘，如果此时n是偶数，直接把上次递归得到的值算个平方返回即可，如果是奇数，则还需要乘上个x的值。还有一点需要引起注意的是n有可能为负数，对于n是负数的情况，我可以先用其绝对值计算出一个结果再取其倒数即可，之前是可以的，但是现在 test case 中加了个负2的31次方后，这就不行了，因为其绝对值超过了整型最大值，会有溢出错误，不过可以用另一种写法只用一个函数，在每次递归中处理n的正负，然后做相应的变换即可，代码如下：

解法一:

class Solution {
public:
    double myPow(double x, int n) {
        if (n == 0) return 1;
        double half = myPow(x, n / 2);
        if (n % 2 == 0) return half * half;
        if (n > 0) return half * half * x;
        return half * half / x;
    }
};

这道题还有迭代的解法，让i初始化为n，然后看i是否是2的倍数，不是的话就让 res 乘以x。然后x乘以自己，i每次循环缩小一半，直到为0停止循环。最后看n的正负，如果为负，返回其倒数，参见代码如下：

解法二：

class Solution {
public:
    double myPow(double x, int n) {
        double res = 1.0;
        for (int i = n; i != 0; i /= 2) {
            if (i % 2 != 0) res *= x;
            x *= x;
        }
        return n < 0 ? 1 / res : res;
    }
};

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