common

全局变量以*号包裹 ------------------------------------------------------------- 列表：(list 元素1 元素2 元素3 ...) 属性表：(list 符号1 元素1 符号2 元素2 ...) 从属性表取值：(getf 属性表 符号) 遍历列表：(dolist 元素 列表)，用PHP表述就相当于foreach(列表 as 元素) 压入元素：(push 元素 变量) ------------------------------------------------------------- 定义函数：(defun 函数名 (参数表) 函数体) 定义变量：(defvar 变量名 变量值) 变量赋值：(setf 变量名 变量值) ------------------------------------------------------------- format格式化参数的含义 # ~{ ... ~}：拆分一个属性表的键值对 # ~a：占位符 # ~t：制表符 # ~%：换行符 来源： oschina 链接： https://my.oschina.net/u/103341/blog/121425

找最长公共前缀，Longest Common Prefix，自己写的小算法，并吸取他人的精华进行改进，直接上代码，第一次写难免会有疏漏，还望大家指正 #include<iostream> #include<string> #include<vector> using namespace std; //第10题：找最长公共前缀，Longest Common Prefix //abcd //ab //abjklg //输出：ab //自己的思路：从第一个串的第一个字符开始匹配，后面的串与第一串进行匹配。如果都能匹配，则第一个串即为最长前缀，若不行，则从不行的那个字符直接推出循环 //原始版本：设置flag和flag1,flag用于确定哪个字符可以加入到最长匹配字符串中，flag1用于当遇到不能匹配的串时来结束外层循环 string longestCommonPrefixpre(vector<string>& strs) { int i,j,flag=0,flag1=0; string com; if(strs.size()==0) //一开始没有通过就是因为没有这行，当strs为空的时候没有考虑进去 return com; for(i=0;i<strs[0].size();i++) { flag=0; for(j=0;j<strs.size();j++) { if(strs[j][i]

235. Lowest Common Ancestor of a Binary Search Tree Total Accepted: 77021 Total Submissions: 205265 Difficulty: Easy Given a binary search tree (BST), find the lowest common ancestor (LCA) of two given nodes in the BST. According to the definition of LCA on Wikipedia : “The lowest common ancestor is defined between two nodes v and w as the lowest node in T that has both v and w as descendants (where we allow a node to be a descendant of itself).” _______6______ / \ ___2__ ___8__ / \ / \ 0 _4 7 9 / \ 3 5 For example, the lowest common ancestor (LCA) of nodes 2 and 8 is 6 . Another example is

https://leetcode.com/problems/lowest-common-ancestor-of-a-binary-search-tree/ Given a binary search tree (BST), find the lowest common ancestor (LCA) of two given nodes in the BST. According to the definition of LCA on Wikipedia : “The lowest common ancestor is defined between two nodes p and q as the lowest node in T that has both p and q as descendants (where we allow a node to be a descendant of itself).” Given binary search tree: root = [6,2,8,0,4,7,9,null,null,3,5] Example 1: Input: root = [6,2,8,0,4,7,9,null,null,3,5], p = 2, q = 8 Output: 6 Explanation: The LCA of nodes 2 and 8 is 6.

题目链接 https://leetcode.com/problems/lowest-common-ancestor-of-a-binary-search-tree/description/ 题目描述 Given a binary search tree (BST), find the lowest common ancestor (LCA) of two given nodes in the BST. According to the definition of LCA on Wikipedia: “The lowest common ancestor is defined between two nodes p and q as the lowest node in T that has both p and q as descendants (where we allow a node to be a descendant of itself ).” Given binary search tree: root = [6,2,8,0,4,7,9,null,null,3,5] _______6______ / \ ___2__ ___8__ / \ / \ 0 _4 7 9 / \ 3 5 Example 1: Input: root = [6,2,8,0,4,7,9,null

https://leetcode.com/problems/lowest-common-ancestor-of-a-binary-search-tree/description/ Given a binary search tree (BST), find the lowest common ancestor (LCA) of two given nodes in the BST. According to the definition of LCA on Wikipedia : “The lowest common ancestor is defined between two nodes v and w as the lowest node in T that has both v and w as descendants (where we allow a node to be a descendant of itself).” _______6______ / \ ___2__ ___8__ / \ / \ 0 _4 7 9 / \ 3 5 For example, the lowest common ancestor (LCA) of nodes 2 and 8 is 6 . Another example is LCA of nodes 2 and 4 is 2 ,

Given a binary search tree (BST), find the lowest common ancestor (LCA) of two given nodes in the BST. According to the definition of LCA on Wikipedia : “The lowest common ancestor is defined between two nodes p and q as the lowest node in T that has both p and q as descendants (where we allow a node to be a descendant of itself).” Given binary search tree: root = [6,2,8,0,4,7,9,null,null,3,5] Example 1: Input: root = [6,2,8,0,4,7,9,null,null,3,5], p = 2, q = 8 Output: 6 Explanation: The LCA of nodes 2 and 8 is 6. Example 2: Input: root = [6,2,8,0,4,7,9,null,null,3,5], p = 2, q = 4 Output: 2

Given a binary search tree (BST), find the lowest common ancestor (LCA) of two given nodes in the BST. According to the definition of LCA on Wikipedia : “The lowest common ancestor is defined between two nodes v and w as the lowest node in T that has both v and w as descendants (where we allow a node to be a descendant of itself).” _______6______ / \ ___2__ ___8__ / \ / \ 0 _4 7 9 / \ 3 5 For example, the lowest common ancestor (LCA) of nodes 2 and 8 is 6 . Another example is LCA of nodes 2 and 4 is 2 , since a node can be a descendant of itself according to the LCA definition. 给定一个二叉搜索树

不规整元素的宽高等比例 在不同屏幕情况中不同宽高的元素都以相同等比例、等宽和等高方式展示。 需求 设计师希望页面的图片区域，以宽高为2:1比例且所有图片的等宽和等高的方式展示。小加同学觉得设计师这需求太容易，分分钟搞定，拿到图片后便开始刷刷的撸代码。原型设计稿大致如下: bootstrap 栅格系统 思路 每个图片区域宽度为父元素宽度的25%，图片的宽度设置100%，其高度根据宽度等比例自动缩放（小加以为图片的宽高应该是同比例的），这样就可以适应屏幕达到要求咯～ HTML <div class="section"> <h1 class="section__title">初版</h1> <div class="section__images row"> <div class="section__image-wrap col-xs-3"> <img class="section__image" src="../../img/common/common-1.jpg"> </div> <div class="section__image-wrap col-xs-3"> <img class="section__image" src="../../img/common/common-2.jpg"> </div> <div class="section__image-wrap col-xs-3

一、题目说明 题目236. Lowest Common Ancestor of a Binary Tree，在一个二叉树中找两个节点的最近公共祖先。难度是Medium！ 二、我的解答 这个用二叉树的递归遍历，稍加改造即可： class Solution{ public: TreeNode* lowestCommonAncestor(TreeNode* root,TreeNode*p,TreeNode*q){ if(root == NULL) return root; if(root == p || root==q) return root; TreeNode* left,*right; left = lowestCommonAncestor(root->left,p,q); right = lowestCommonAncestor(root->right,p,q); if(left !=NULL && right!=NULL){ return root; }else if(left != NULL){ return left; }else if(right != NULL){ return right; }else{ return NULL; } } }; 性能如下： Runtime: 16 ms, faster than 94.88% of C++ online