陪suki去coffee chat丢烘，出去走了走，還上了willis tower些椒，然而下雨播瞳，全都被云擋住了，好不容易有點(diǎn)風(fēng)免糕，把云撥開了赢乓，剛想擺個(gè)pose，另一片云又來了石窑。牌芋。。
Reverse String
沒什么好說的松逊，兩個(gè)指針Reverse Words in a String
I love yahoo --> yahoo love I
先reverse一下整個(gè)String
然后再以空格為界躺屁，用兩個(gè)指針確定單一word，reverse單一word
Right Shift By N Characters
abcdef k = 4 -> abcd | ef -> dcba | fe -> ef | abcd
同上題经宏，先reverse拆分部分犀暑，再reverse所有
type5 char replacement
ex. "student" -> "stuxxt" (den -> xx) s1 to s2
two pointer
case1: s1.length() >= s2.length()
照常兩個(gè)pointer來做
case2：s1.length() < s2.length() (string的長度會(huì)超出原來的)
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（要考慮奇偶性），是的保證chunk1.size = chunk3.size, chunk2.size = chunk4.size烁兰，將中間兩部分reverse耐亏，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
兩個(gè)pointer做， 加個(gè)count缚柏，第一遍做case1并保留單個(gè)出現(xiàn)的字符 變成a4bc2a5苹熏，第二遍做case2碟贾，倒著做得出結(jié)果币喧，類似前面的char replacement
Sliding window
也是2pointer轨域，但是物理意義不同，[slow杀餐， fast]中間才是需要的干发，slow前面和fast后面沒用
Longest Substring Without Repeating Characters
為保證沒有重復(fù)char，先用hashset記錄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史翘，# of char already matched
需要理清思路枉长，有必要重新寫一遍
今天看的string2 的lec，各個(gè)例題實(shí)在是有難度琼讽，都是onsite級(jí)別的題必峰，思路理解差不多了，但是implement起來還是困難钻蹬，先放一放
