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<n; ++i) {
            for (int j=0; j<read.size(); ++j ) {
                if (read[j]!=last) {
                    if (last!='\0') {
                        write.push_back(ct+'0');
                        write.push_back(last);
                    }
                    last = read[j];
                    ct = 1;
                } else {
                    ++ct;
                }
            }
            write.push_back(ct+'0');
            write.push_back(last);
            read = write;
            write = "";
            last = '\0';
            ct = 0;
        }
        return read;
    }

2. 内置STL

auto nexti = find_if(i, curr.end(), bind1st(not_equal_to<char>(), *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<GIVE TYPE HERE>(_, _)
• stringstream
  used in original answer, can be used to convert different types into string, eg. ss<<distance(i, nexti)<<*i;

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<curr.end();) {
            auto nexti = find_if(i, curr.end(), bind1st(not_equal_to<char>(), *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<s.size(); ++i) {
            ++record[s[i]-'a'];
            --record[t[i]-'a'];
        }
        for(int i=0; i<24; ++i) {
            if (record[i]) return false;
        }
        return true;
    }

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<string> record;
        for (auto i=path.begin(); i<path.end();) {
            ++i;
            auto lenCurr = find(i, path.end(), '/');
            string curr = string(i, lenCurr); 
            
            if (curr !="." && curr !="") {
                if (curr=="..") {
                    if (!record.empty()) {
                        record.pop();
                    }
                } else {
                    record.push(curr);
                }
            }
            i = lenCurr;
        }
        string ret;
        while (!record.empty()) {
            ret.insert( 0, record.top()); 
            ret.insert( 0, "/");
            record.pop();
        }
        return ret==""? "/" : ret;
    }

4] Length of Last Word

注意去词尾空格就好

    int lengthOfLastWord(string s) {
        auto i=s.rbegin(); 
        while (i<s.rend() && *i==' ') ++i; 
        
        int len = 0;
        for (; i<s.rend(); ++i) {
            if (*i==' ') break;
            else ++len;
        }
        return len;
    }