Implement pow(x, n). 幂运算,简单的方法snag然很好实现,直接循环相乘就可以了,但是这里应该不是那种那么简单,我的做法使用到了一点递归: class Solution { public: double myPow(double x, int n) { if(n == INT_MIN) return 1.0/(x * myPow(x, INT_MAX)); //这里考虑了一下由于传参数的限制还是将其变成INT_MAX ) return 1.0/myPow(x, -n); )…

博客搬至blog.csgrandeur.com,cnblogs不再更新. 新的题解会更新在新博客:http://blog.csgrandeur.com/2014/01/15/LeetCode-OJ-Solution/ ———————————————————————————————————————— ———————————————————————————————————————— LeetCode OJ 题解 LeetCode OJ is a platform for preparing tech…

leetcode 326. Power of Three(不用循环或递归) Given an integer, write a function to determine if it is a power of three. Follow up: Could you do it without using any loop / recursion? 题意是判断一个数是否是3的幂,最简单的也最容易想到的办法就是递归判断,或者循环除. 有另一种方法就是,求log以3为底n的对数.类似 如果n=9,则…

Problem Link: http://oj.leetcode.com/problems/interleaving-string/ Given s1, s2, s3, find whether s3 is formed by the interleaving of s1 and s2. For example, Given: s1 = "aabcc", s2 = "dbbca", When s3 = "aadbbcbcac", return t…

Problem link: http://oj.leetcode.com/problems/reverse-words-in-a-string/ Given an input string, reverse the string word by word. For example, Given s = "the sky is blue", return "blue is sky the". LeetCode OJ supports Python now! The s…

一直没有系统地学习过算法,不过算法确实是需要系统学习的.大二上学期,在导师的建议下开始学习数据结构,零零散散的一学期,有了链表.栈.队列.树.图等的概念.又看了下那几个经典的算法——贪心算法.分治算法.动态规划以及回溯算法.不过,都是知其一不知其二的一知半解.到最后,发现学到了一堆的一知半解,在学校课程开这门课时,却又是不得不再学一遍.学一样东西,又不全心全意地主动去把它学好,到最后只是花费了时间,却没真真学到东西.所以,不求学识有多广,但求所学的都精. 一个偶然的机会,让我在网上找到了一本社区…

Serialization is the process of converting a data structure or object into a sequence of bits so that it can be stored in a file or memory buffer, or transmitted across a network connection link to be reconstructed later in the same or another comput…

常泡LC的朋友知道LC是不提供代码打包下载的,不像一般的OJ,可是我不备份代码就感觉不舒服- 其实我想说的是- 我自己写了抓取个人提交代码的小工具,放在GitCafe上了- 不知道大家有没有兴趣 https://gitcafe.com/aaronzhou/ ... ssion 拿Java写的,也打了一个包,本地有JRE环境就能运行,欢迎拍砖 抓取 LeetCode OJ 个人提交的代码 Gradle 构建项目 HttpClient.jsoup 抓取.解析网页 使用说明 进入release文件夹…

题目: Given a 2D binary matrix filled with 0's and 1's, find the largest square containing all 1's and return its area. For example, given the following matrix: 1 0 1 0 0 1 0 1 1 1 1 1 0 0 1 0 Return 4. 题目链接:https://leetcode.com/problems/maximal-square…

Given an integer, convert it to a roman numeral. Input is guaranteed to be within the range from 1 to 3999. 给定一个整数,将他转换到罗马字符.同样这里的罗马字符不会超过1000(M),PS:关于罗马字符的介绍看我的这篇文章 :LeetCode OJ:Roman to Integer(转换罗马字符到整数) 这里的大体思路是这样的,例如对于罗马字符952来说,起答题的可以分成三个组成部分, 就…

Serialization is the process of converting a data structure or object into a sequence of bits so that it can be stored in a file or memory buffer, or transmitted across a network connection link to be reconstructed later in the same or another comput…

Given an integer, write a function to determine if it is a power of two. 2的幂的二进制表示中,必定仅仅有一个"1",且不可能为负数. class Solution { public: bool isPowerOfTwo(int n) { if(n<0) {//若为负数则直接返回 return false; } int num=0; while(n) {//统计1的个数 n=n&(n-1); num+…

Given an integer, write a function to determine if it is a power of two. Example 1: Input: 1 Output: true Example 2: Input: 16 Output: true Example 3: Input: 218 Output: false 给一个整数,写一个函数来判断它是否为2的次方数. 利用计算机用的是二进制的特点,用位操作,此题变得很简单. 2的n次方的特点是:二进制表示中最高位是…

Given an integer (signed 32 bits), write a function to check whether it is a power of 4. Example:Given num = 16, return true. Given num = 5, return false. Follow up: Could you solve it without loops/recursion? Credits:Special thanks to @yukuairoyfor…

[九度OJ]题目1474:矩阵幂 解题报告 标签(空格分隔): 九度OJ http://ac.jobdu.com/problem.php?pid=1474 题目描述: 给定一个n*n的矩阵,求该矩阵的k次幂,即P^k. 输入: 输入包含多组测试数据. 数据的第一行为一个整数T(0<T<=10),表示要求矩阵的个数. 接下来有T组测试数据,每组数据格式如下: 第一行:两个整数n(2<=n<=10).k(1<=k<=5), 两个数字之间用一个空格隔开,含义如上所示. 接下来…

Given an integer, write a function to determine if it is a power of two. 看一个数是不是2的幂,代码如下: class Solution { public: bool isPowerOfTwo(int n) { ) return false; bool isPower = false; ; i < n; i<<=){ if(!isPower && (i&n)) isPower = true;…

描述 Given an integer, write a function to determine if it is a power of two. 给定一个整数,编写一个函数来判断它是否是 2 的幂次方. 解析 2的幂只有1个1 仔细观察,可以看出 2 的次方数都只有一个 1 ,剩下的都是 0 .根据这个特点,只需要每次判断最低位是否为 1 ,然后向右移位,最后统计 1 的个数即可判断是否是 2 的次方数. 减一法 如果一个数是 2 的次方数的话,那么它的二进数必然是最高位为1,其它都为 0…

翻译 给定一个整型数,写一个函数决定它是否是3的幂(翻译可能不太合适-- 跟进: 你能否够不用不论什么循环或递归来完毕. 原文 Given an integer, write a function to determine if it is a power of three. Follow up: Could you do it without using any loop / recursion? 分析 题意我事实上不是满懂,比方说12究竟可不能够呢?还是说仅仅有:3.9.27.81这样的才行…

1.题目:原题链接 Given an integer, write a function to determine if it is a power of two. 给定一个整数,判断该整数是否是2的n次幂. 2.思路 如果一个整数是2的n次幂,那么首先其应当是正数,其次该数的二进制表示必定是以1开头,后续若有数字必为0. 3.代码 class Solution { public: bool isPowerOfTwo(int n) { return (!(n&(n-1)))&&n&…

326. Power of Three Question Total Accepted: 1159 Total Submissions: 3275 Difficulty: Easy 推断给定整数是否是3的某次方. Given an integer, write a function to determine if it is a power of three. Follow up: Could you do it without using any loop / recursion? 完毕此题.…

题意:判断1个数n是否刚好是2的幂,幂大于0. 思路:注意会给负数,奇数.对于每个数判断31次即可. class Solution { public: bool isPowerOfTwo(int n) { ||(n&)==&&n>) return false; unsigned ; while(t<n) //注意爆int t<<=; if(t==n) return true; return false; } }; AC代码 一行代码,更巧的写法: 如果一个数…

/* 用位操作,乘2相当于左移1位,所以2的幂只有最高位是1 所以问题就是判断你是不是只有最高位是1,怎判断呢 这些数-1后形成的数,除了最高位,后边都是1,如果n&n-1就可以判断了 如果是2的幂,&的结果是全0 */ if (n<=0) return false; return ((n&(n-1))==0); 划重点: 一个数*2,出相当于左移一位 2.判断是不是3的幂,没啥用的一道题 public boolean isPowerOfThree(int n) { /* 不…

Valid Palindrome吐槽一下Leetcode上各种不定义标准的输入输出(只是面试时起码能够问一下输入输出格式...),此篇文章不是详细的题解,是自己刷LeetCode的一个笔记吧,尽管没有代码,可是略微难一点的都会标出主要思路,便于以后复习 PS:题目中有"水题"两字的都是一遍AC,没有标注的都说明了问题,顺序依照Leetcode上时间倒序排列,少量题目因为和之前的题目有相关性,因此将其放在一起,比方12题和13题,因此中间可能会"缺少"几道题目,缺少的…

这三道题目都是一个意思,就是判断一个数是否为2/3/4的幂,这几道题里面有通用的方法,也有各自的方法,我会分别讨论讨论. 原题地址:231 Power of Two:https://leetcode.com/problems/power-of-two/description/ 326 Power of Three:https://leetcode.com/problems/power-of-three/description/ 342 Power of Four :https://leetcod…

Problem Link: https://oj.leetcode.com/problems/validate-binary-search-tree/ We inorder-traverse the tree, and for each node we check if current_node.val > prev_node.val. The code is as follows. # Definition for a binary tree node # class TreeNode: #…

Problem Link: https://oj.leetcode.com/problems/recover-binary-search-tree/ We know that the inorder traversal of a binary search tree should be a sorted array. Therefore, we can compare each node with its previous node in the inorder to find the two…

Problem Link: https://oj.leetcode.com/problems/same-tree/ The following recursive version is accepted but the iterative one is not accepted... # Definition for a binary tree node # class TreeNode: # def __init__(self, x): # self.val = x # self.left =…

Problem Link: https://oj.leetcode.com/problems/symmetric-tree/ To solve the problem, we can traverse the tree level by level. For each level, we construct an array of values of the length 2^depth, and check if this array is symmetric. The tree is sym…

Problem Link: https://oj.leetcode.com/problems/binary-tree-level-order-traversal/ Traverse the tree level by level using BFS method. # Definition for a binary tree node # class TreeNode: # def __init__(self, x): # self.val = x # self.left = None # se…

Problem Link: https://oj.leetcode.com/problems/binary-tree-zigzag-level-order-traversal/ Just BFS from the root and for each level insert a list of values into the result. # Definition for a binary tree node # class TreeNode: # def __init__(self, x):…