这个题目我也没有思路,同学们可以查看这个http://www.cnblogs.com/NickyYe/p/4442867.html 下面是我改进后的代码 第一种方法: class Solution { public: double myPow(double x, int n) { == n); ); )) return half*half; ) { return half*half*x; } else { / x); } } }; 第二种方法: == n); double result = 1.…

前言   [LeetCode 题解]系列传送门:  http://www.cnblogs.com/double-win/category/573499.html 1.题目描述 Implement pow(x, n). 2. 思路 题目很精简,不过需要考虑的情况很多,特别的需要注意 n可能为0 或者负数的情况. 另外,通过二分法,可以减少计算量. 3. 解法 class Solution { public: double pow(double x,int n) { ==x || ) return…

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:…

Implement pow(x, n). 解题思路: 直接使用乘法实现即可,注意下,如果n很大的话,递归次数会太多,因此在n=10和n=-10的地方设置一个检查点,JAVA实现如下: static public double myPow(double x, int n) { if(n==1) return x; else if(n>1&&n<=10) return myPow(x,n-1)*x; if(n>10) return myPow(myPow(x,10),n/10…

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 题意: 求…

Pow(x, n) Implement pow(x, n). 思路:题目不算难.可是须要考虑的情况比較多. 详细代码例如以下: public class Solution { public double myPow(double x, int n) { boolean isMin0 = true;//结果负号 if(x > 0 || (n&1) == 0){//x>0或n为偶数 isMin0 = false;//为正 } x = x < 0 ? -x:x;//将x统一设为正值 d…

Pow(x, n) Total Accepted: 25273 Total Submissions: 97470My Submissions Implement pow(x, n). 题意:求x的n次幂 思路:二分法 n有可能是负的或正的 当n为负是,pow(x, n) = 1/pow(x, -n) x^n = x^{n/2} * x^{n/2}* x^{n%2} 复杂度:时间O(log n).空间O(1) double power(double x, int n){ if(n == 0) re…

Your task is to calculate ab mod 1337 where a is a positive integer and b is an extremely large positive integer given in the form of an array. Example1: a = 2 b = [3] Result: 8 Example2: a = 2 b = [1,0] Result: 1024 Credits:Special thanks to @Stomac…

Implement the following operations of a queue using stacks. push(x) -- Push element x to the back of queue. pop() -- Removes the element from in front of queue. peek() -- Get the front element. empty() -- Return whether the queue is empty. Notes: You…

Implement the following operations of a stack using queues. push(x) -- Push element x onto stack. pop() -- Removes the element on top of the stack. top() -- Get the top element. empty() -- Return whether the stack is empty. Notes: You may assume that…