2018-3 reverse words

Posted 2020-10-20 hisonsandiego

亚麻的OA题

reverse words（leetcode有），要求尽可能time／space efficient

 

1，input的string的前后空格都要去除

2，若是input string 中的俩个word 之间有多个空格， 则仅需要保留一个。

 

Clarification:

 

• What constitutes a word?
  A sequence of non-space characters constitutes a word.
• Could the input string contain leading or trailing spaces?
  Yes. However, your reversed string should not contain leading or trailing spaces.
• How about multiple spaces between two words?
  Reduce them to a single space in the reversed string.

 

class Solution {
public:
    void reverseWords(string &s) {
        if (s.empty()) return ;
        if ( s.size () == 1 && s !=" ") return ; // only one char 
 
        
        //cout <<" the size of the input string is large than 1"<< endl;
        int st =0;
        int ed = s.size ()-1;
        int size = s.size();
        string s2;
        
        while ( st < size  && s[st] == ‘ ‘) st++;
        if ( st == size ) { s=""; return ;}
        
        while ( ed >0 && s[ed] == ‘ ‘) ed--;
        
        int p1  = ed;
        int p2 = p1;
        cout << " all the sapce is removed, and size is , "<< size << " st is "<< st<<" end is " << ed<<endl;
        while ( p1 >= st  ){
            if (s[p1] == ‘ ‘){ p1--; continue ;}
        
            
            p2 = p1;
            while ( p2 >= st ){
              //  cout <<"p2 is " << p2 << " p1 is " << p1 <<endl;
                if ( p2 == st) {//the last char.
               //    cout <<" the last char"<< endl;
                    s2 += s.substr(p2, p1 -p2 +1);
                    s = s2;
                    return ;
                }
                
                if ( s[p2] != ‘ ‘) p2 --;
                else{
                     s2 += s.substr(p2+1, p1-p2);
             //        cout << " copy the word "<< s.substr(p2+1, p1-p2) <<" from "<< p2 << " to "<< p1<< endl;;
                     s2 += ‘ ‘;
                     p1 = p2;
                    break;
                } 
            }
        }
        //s = s2;
        return ;
    }
};