7月缺米的来刷题/Mock interview活动

Day 10
Review

今天刷了5题：
Minimum Height Trees
Best Time to Buy and Sell Stock with Cooldown
Factor Combinations
Maximum Number of Non-Overlapping Substrings
Remove Linked List Elements

7/20 打卡 day6
用having + aggregate function 必用group by
join 之前可以先把其中一个table精简总结

7/20 tree

day 17

今天mock又是2sum。。hmmm

lc 10+
Design Search Autocomplete System   0 minutes ago
AcceptedjavaDesign Search Autocomplete System   1 minute ago
AcceptedjavaDesign Search Autocomplete System   1 minute ago
Wrong AnswerjavaDesign Search Autocomplete System   7 minutes ago
AcceptedjavaDesign Search Autocomplete System   8 minutes ago
Wrong AnswerjavaDesign Search Autocomplete System   8 minutes ago
AcceptedjavaStickers to Spell Word   1 hour, 1 minute ago
AcceptedjavaStickers to Spell Word   1 hour, 5 minutes ago
AcceptedjavaStickers to Spell Word   1 hour, 42 minutes ago
AcceptedjavaCount Different Palindromic Subsequences   5 hours, 5 minutes ago
AcceptedjavaCount Different Palindromic Subsequences   5 hours, 17 minutes ago
AcceptedjavaCount Different Palindromic Subsequences   5 hours, 21 minutes ago
AcceptedjavaCount Different Palindromic Subsequences   5 hours, 21 minutes ago
AcceptedjavaCount Different Palindromic Subsequences   5 hours, 50 minutes ago
AcceptedjavaNumber of Closed Islands   7 hours, 50 minutes ago
AcceptedjavaNumber of Closed Islands   7 hours, 53 minutes ago
AcceptedjavaNumber of Closed Islands   8 hours, 16 minutes ago
AcceptedjavaContainer With Most Water   8 hours, 31 minutes ago
Compile ErrorjavaContainer With Most Water   8 hours, 31 minutes ago
AcceptedjavaContainer With Most Water   8 hours, 40 minutes ago

7-21，有点阅读理解了变成。

20200717：
3： 0256, 0276, 0487
4： 0076, 0079, 0265, 1289
5：

7.21打卡，4道medium和4道easy

日期: 7/21/2020
题目: [2]
79. Word Search
207. Course Schedule

谢谢各位大佬给分，继续努力
D19：  打卡第19天 (UTC: 21/07/2020)
继续学习 dp
1： Jump Game II （45）

状态：dp[i]从起点出发跳到i点需要步数
转移方程：dp[i] = min(dp[j]+1, dp[i]),j满足条件可以一步跳到i，加个判断
初始化：dp[0] = 0
答案：dp[nums.length - 1 ]，最后一个元素状态

Time Limit Exceeded

贪心法：

实时维护最远可达的位置rightmost，和上次起跳点能到达的最远边界end。step是跳跃次数，即最终的返回结

2：Longest Increasing Subsequence （300）

维护一个一维 dp 数组，其中 dp[i] 表示以 nums[i] 为结尾的最长递增子串的长.
对于每一个 nums[i]，从第一个数再搜索到i，如果发现某个数小于 nums[i]，更新 dp[i]，更新方法为 dp[i] = max(dp[i], dp[j] + 1)，
即比较当前 dp[i] 的值和那个小于 num[i] 的数的 dp 值加1的大小，就这样不断的更新 dp 数组，到最后 dp 数组中最大的值就是 LIS 的长度