6.28 ctAndSay & anagrams & simplifyPath & lenOfLastWord

注重medium先不做hard吧

- to do

anagram写过了

1] Count and Say

** 1.naive的模拟，73%**

    string countAndSay(int n) {
        if (n<2) return "1";
        
        string read ="1", write = "";
        char last = '\0';
        int ct = 0;
        for (int i=1; i

2. 内置STL

auto nexti = find_if(i, curr.end(), bind1st(not_equal_to(), *i));

• findif (iterator start, iterator end, unary operation) :
  within [first, end),
  -> return the iterator pointing to the first elem that returns T under the unary operation,
  -> else return end
• bind1st/bind2nd(const Operation& op, const T& x)
• not_equal_to(_, _)
• stringstream
  used in original answer, can be used to convert different types into string, eg. ss<

istringstream类是从istream和stringstreambase派生而来, 以此类推见图

    void getNext (string& curr) {
        string ss;
        for (auto i=curr.begin(); i(), *i));
            ss+= '0'+ distance(i, nexti);
            ss+= *i;
            i = nexti;
        }
        swap(curr, ss);
    }
    
    string countAndSay(int n) {
        string curr = "1";
        while (--n) getNext(curr);
        return curr;
    }

2] Valid Anagram

基本最快的

用 int record[24] = {somenum}; 来确保initialization，内构是第一个设为somenum，其余0

    bool isAnagram(string s, string t) {
        if (s.length()!=t.length()) return false;
        int record[24] = {0};
        for (int i=0; i

3】 Simplify Path

自己写没考虑全，mark重写
中间loop可以精简但是考虑到会增加时间。

• find (iterator begin, iterator end, const char)
  returns iterator pointing to found one
• distinct from someStr.find (char c OR string& OR char*, size_t pos=0, size_type n)
  pos (op): starting to search from here, default to be 0 if not specified
  n (op): len of sequence of chars (needle) to match
• also += seems work faster then append

    string simplifyPath(string path) {
        stack record;
        for (auto i=path.begin(); i

4] Length of Last Word

注意去词尾空格就好

    int lengthOfLastWord(string s) {
        auto i=s.rbegin(); 
        while (i