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;
    }
};

到此这篇关于C++实现LeetCode(50.求x的n次方)的文章就介绍到这了,更多相关C++实现求x的n次方内容请搜索我们以前的文章或继续浏览下面的相关文章希望大家以后多多支持我们！

赞 (0)

相关推荐

  • C++实现LeetCode(144.二叉树的先序遍历)

    [LeetCode] 144. Binary Tree Preorder Traversal 二叉树的先序遍历 Given a binary tree, return the preorder traversal of its nodes' values. Example: Input:  [1,null,2,3] 1 \ 2 / 3 Output:  [1,2,3] Follow up: Recursive solution is trivial, could you do it iterat

  • C++实现LeetCode(78.子集合)

    [LeetCode] 78. Subsets 子集合 Given a set of distinct integers, S, return all possible subsets. Note: Elements in a subset must be in non-descending order. The solution set must not contain duplicate subsets. For example, If S = [1,2,3], a solution is:

  • C++实现LeetCode(41.首个缺失的正数)

    [LeetCode] 41. First Missing Positive 首个缺失的正数 Given an unsorted integer array, find the smallest missing positive integer. Example 1: Input: [1,2,0] Output: 3 Example 2: Input: [3,4,-1,1] Output: 2 Example 3: Input: [7,8,9,11,12] Output: 1 Note: Your

  • C++实现LeetCode(113.二叉树路径之和之二)

    [LeetCode] 113. Path Sum II 二叉树路径之和之二 Given a binary tree and a sum, find all root-to-leaf paths where each path's sum equals the given sum. For example: Given the below binary tree and sum = 22,  5 / \ 4   8 /      / \ 11  13  4 /  \         / \ 7  

  • C++实现LeetCode(47.全排列之二)

    [LeetCode] 47. Permutations II 全排列之二 Given a collection of numbers that might contain duplicates, return all possible unique permutations. Example: Input: [1,1,2] Output: [ [1,1,2], [1,2,1], [2,1,1] ] 这道题是之前那道 Permutations 的延伸,由于输入数组有可能出现重复数字,如果按照之前的

  • C++实现LeetCode(94.二叉树的中序遍历)

    [LeetCode] 94. Binary Tree Inorder Traversal 二叉树的中序遍历 Given a binary tree, return the inorder traversal of its nodes' values. Example: Input: [1,null,2,3] 1 \ 2 / 3 Output: [1,3,2] Follow up: Recursive solution is trivial, could you do it iteratively

  • C++实现LeetCode(90.子集合之二)

    [LeetCode] 90. Subsets II 子集合之二 Given a collection of integers that might contain duplicates, S, return all possible subsets. Note: Elements in a subset must be in non-descending order. The solution set must not contain duplicate subsets. For example

  • C++实现LeetCode(40.组合之和之二)

    [LeetCode] 40. Combination Sum II 组合之和之二 Given a collection of candidate numbers (candidates) and a target number (target), find all unique combinations in candidates where the candidate numbers sums to target. Each number in candidates may only be u

  • 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

  • C++实现LeetCode(58.求末尾单词的长度)

    [LeetCode] 58. Length of Last Word 求末尾单词的长度 Given a string s consists of upper/lower-case alphabets and empty space characters ' ', return the length of last word in the string. If the last word does not exist, return 0. Note: A word is defined as a

  • C++实现LeetCode( 69.求平方根)

    [LeetCode] 69. Sqrt(x) 求平方根 Implement int sqrt(int x). Compute and return the square root of x, where x is guaranteed to be a non-negative integer. Since the return type is an integer, the decimal digits are truncated and only the integer part of the

  • C++实现LeetCode(187.求重复的DNA序列)

    [LeetCode] 187. Repeated DNA Sequences 求重复的DNA序列 All DNA is composed of a series of nucleotides abbreviated as A, C, G, and T, for example: "ACGAATTCCG". When studying DNA, it is sometimes useful to identify repeated sequences within the DNA. Wr

  • C++实现LeetCode(124.求二叉树的最大路径和)

    [LeetCode] 124. Binary Tree Maximum Path Sum 求二叉树的最大路径和 Given a non-empty binary tree, find the maximum path sum. For this problem, a path is defined as any sequence of nodes from some starting node to any node in the tree along the parent-child conn

  • C++实现LeetCode(129.求根到叶节点数字之和)

    [LeetCode] 129. Sum Root to Leaf Numbers 求根到叶节点数字之和 Given a binary tree containing digits from 0-9 only, each root-to-leaf path could represent a number. An example is the root-to-leaf path 1->2->3 which represents the number 123. Find the total sum

  • C++实现LeetCode(128.求最长连续序列)

    [LeetCode] 128.Longest Consecutive Sequence 求最长连续序列 Given an unsorted array of integers, find the length of the longest consecutive elements sequence. Your algorithm should run in O(n) complexity. Example: Input: [100, 4, 200, 1, 3, 2] Output: 4 Expl

  • C++实现LeetCode(152.求最大子数组乘积)

    [LeetCode] 152. Maximum Product Subarray 求最大子数组乘积 Given an integer array nums, find the contiguous subarray within an array (containing at least one number) which has the largest product. Example 1: Input: [2,3,-2,4] Output: 6 Explanation: [2,3] has

  • C++实现LeetCode(160.求两个链表的交点)

    [LeetCode] 160.Intersection of Two Linked Lists 求两个链表的交点 Write a program to find the node at which the intersection of two singly linked lists begins. For example, the following two linked lists: A:          a1 → a2 c1 → c2 → c3             B:     b1

  • C++实现LeetCode(162.求数组的局部峰值)

    [LeetCode] 162.Find Peak Element 求数组的局部峰值 A peak element is an element that is greater than its neighbors. Given an input array nums, where nums[i] ≠ nums[i+1], find a peak element and return its index. The array may contain multiple peaks, in that c

随机推荐