8月刷题/Mock interview接龙活动（缺米刷题的来）

刷了3题：
Reorganize String
Largest Plus Sign
Iterator for Combination

20200808:
new
3: 1286
4:
5:

review
3: 0039, 0040, 0046, 0047, 0056, 0133, 0938, 1498
4: 0023, 0124, 0138, 0269, 0545
5: 0301

D7 打卡： implement trie

Day 13

1286. Iterator for Combination, 如何用bit map 生成所有的combination
400. Nth Digit, 挺不错的数学题

刷题打卡，8道题
Nth Digit 数数字的个数
Elimination Game 用两个函数相互递归
Rotate Function 这个找通项公式 Fk - Fk-1
Perfect Rectangle 数重复overlap的顶点，要注意也要比较面积，因为会有完全一模一样的input rectangule
Binary Watch 可以用backtrack，但是直接按时间循环更快

8.13 打卡第11天

最近各种准备oa刷题都少了
两道带distance的bfs，都可以用dp做
飞行棋注意：可以有多个src-》des，所以用map<des, list<src>>
src -> des步数耗费0

8.14第14天
一道hard又debug了一天。。
矩阵里找连续最长递增序列。用记忆化搜索。

8月第11天打卡～

第二天打卡
加上我对first missing positive的notes
O(n) time and O(1) space
        we first find if 1 exists in this list, if it doesn't, simply return 1
        when we are iterating the first time, we can turn all the numbers <= 0 and > n to 1
        since that we know if 1 exists, setting these numbers to 1 will be safe
        reason we still set numbers > n to 1 is because this list can only contain 1...n elements
        which means if a number is > n, our return integer will be in 1...n by pidgeon hole principle
        
        in the next loop, we negate the nums[value] to indicate that we have seen this number
        remember to use abs(value)-1 because
        1. the current value can be negative because some other number has negated this value
        2. -1 because array starts from 0 to n-1 but our values are from 1 to n
        
        in the last iteration, we set the starting point to 2, and see if any point from low-1
        is negative. if so, return low
        we will return low outside of the while loop in case 2 or n+1 is the answer

打卡第40天 2020.08.14
669. 修剪二叉搜索树
700. 二叉搜索树中的搜索
671. 二叉树中第二小的节点
872. 叶子相似的树
897. 递增顺序查找树
687. 最长同值路径
965. 单值二叉树
993. 二叉树的堂兄弟节点