缺大米的请参加这个刷题/Mock interview活动

6.22 day23
127 medium 加上一些辅助函数就是最基础的bfs，生成nextWord的方式，需要根据数据规模的变化来决定
126 hard bfs+dfs 先bfs确定最短长度，再逆向dfs找最短路径
133 medium 又是万能的hashmap

打卡第17天
06/22
笔记在这边

https://www.1point3acres.com/bbs/thread-635271-1-1.html
(出处: 一亩三分地)

• Missing number (unsorted)
• Single Number II / Single Number
• Single Number III
• Binary Number with Alternating Bits

Day 14
Today I did a mock interview with a friend, here are some of the advices he gave me:
1. improve overall problem solving speed by 25% - this can be achieved by ensuring I can write brute force dfs/bfs/topo solution in 5 - 10 minutes to save the time for optimizing the solution.
2. Make sure I can answer time complexity related questions within 30 seconds - I think this can be achieved by doing time complexity analysis practices.
3. Do more graph problems and hashtable manipulation problems, make myself comfortable with using hashtables to reduce runtime.

做了5道binary search的题目，有两道简单的，算是复习和巩固一下自己的知识点吧

6.22 打卡三题

总共第13天，今天被问到一个binarySearch题，一开始没想到，但给出了一个O(n)的解，面试官要提高，其实这里我就应该立刻想到用binarysearch了，因为只能是O（logn）, 但我一下子没想好怎么实现binarySearch, 就转向胡思乱想别的解法了。这个经验还挺有价值的，O（n)还要提高就立刻往binarySearch想。

06.21 打卡第十天 刷题第十天
第十天未被加米了，楼主救我！

Leetcode：
724. Find Pivot Index

Mindset: Prefix sum
find the sum first then linear scan from leftMost to rightMost again
compare with two sums and get a conclusion.

561. Array Partition I

Mindset: sort it first then we can sum up all the even(index) number;

Binary Search

3.Classical Binary Search
Use two pointers
always maintain a size >= 2 searching range
then post-processing

4.First Occurance
if (array[left] <= target) {
  left = mid;
}
the left pointer does not jump through mid
Right pointer jump one for the right;

5.Last Occurance
similarliy:
if (array[right] >= target) {
  right = mid;
}
the right pointer does not jump through mid
left pointer jump one for the left;

6. Find the Floor/Ceiling in a sorted array

watch out for the corner case, which is all the element is larger than target, so can not find Floor, return -1;
watch out for the corner case, which is all the element is smaller than target, so can not find Ceiling, return -1;

7.Search in sorted unknown sized array

Handle the right side of the case, watch out for the mid or right would point to null;

第一天打卡，刷了三道题。都是动态规划的内容，现在已经比较有经验了，主要都分为三步：1. 定义 dp 数组；2. 找到 base case；3. 找到状态转移方程。

day 11 #sql
1. explain 查看执行计划
2. 创建表default()
3. insert into values (), ()/ insert or ignore into...如果存在即忽略
4. 创建新表并导入其他表数据create table... (as) select... from...

打卡第八天刷了3个题，1个easy1个medium1个hard
（版主前七天都没加米，第一天在1053楼，第二天在1111楼，第三天在1166，第四天1234，第五天1286，第六天1327，第七天1385， 求加米！！QAQ）