陪suki去coffee chat，出去走了走，还上了willis tower，然而下雨，全都被云挡住了，好不容易有点风，把云拨开了，刚想摆个pose，另一片云又来了。。。
Reverse String
没什么好说的，两个指针Reverse Words in a String
I love yahoo --> yahoo love I
先reverse一下整个String
然后再以空格为界，用两个指针确定单一word，reverse单一word
Right Shift By N Characters
abcdef k = 4 -> abcd | ef -> dcba | fe -> ef | abcd
同上题，先reverse拆分部分，再reverse所有
type5 char replacement
ex. "student" -> "stuxxt" (den -> xx) s1 to s2
two pointer
case1: s1.length() >= s2.length()
照常两个pointer来做
case2：s1.length() < s2.length() (string的长度会超出原来的)
step1: compute the new length after replacement
step2: use two pointer, moving from right to left to guarantee slow is always behind fast.
String shuffling
ABCDE12345 -> A1B2C3D4E5
分成四份（要考虑奇偶性），是的保证chunk1.size = chunk3.size, chunk2.size = chunk4.size，将中间两部分reverse，最后再merge
String permutation
has duplicate letters
String en/decoding
aaaabccaaaaa -> a4b1c2a5
case1: unchaged/shorter -> # of occ >= 2
case2: longer -> # of occ < 2
两个pointer做， 加个count，第一遍做case1并保留单个出现的字符 变成a4bc2a5，第二遍做case2，倒着做得出结果，类似前面的char replacement
Sliding window
也是2pointer，但是物理意义不同，[slow， fast]中间才是需要的，slow前面和fast后面没用
Longest Substring Without Repeating Characters
为保证没有重复char，先用hashset记录the chars in the sliding window
if str[fast] not in hashset, add it and fast++
if str[fast] in hashset, remove str[slow], slow++Find All Anagrams in a String
fixed length
hashmap to record the count of distinct chars
record matchChars，# of char already matched
需要理清思路，有必要重新写一遍
今天看的string2 的lec，各个例题实在是有难度，都是onsite级别的题，思路理解差不多了，但是implement起来还是困难，先放一放
