Leetcode daily exercise: sort sentences

1859. Sort the sentences

The difficulty is simple 15

One The sentence It refers to a sequence of words connected by a single space , And there are no spaces at the beginning and end . Every word contains only lowercase or uppercase letters .

We can add... To a sentence from 1 Start word position index , And put all the words in the sentence Out of order .

• For example , The sentence "This is a sentence" Can be scrambled to get "sentence4 a3 is2 This1" perhaps "is2 sentence4 This1 a3" .

To give you one Out of order Sentences s , It contains no more than 9 individual , Please reconstruct and get the sentences in the original order .

Example 1：

 Input ：s = "is2 sentence4 This1 a3"
 Output ："This is a sentence"
 explain ： take  s  The words in are sorted according to their initial positions , obtain  "This1 is2 a3 sentence4" , Then delete the number .

Example 2：

 Input ：s = "Myself2 Me1 I4 and3"
 Output ："Me Myself and I"
 explain ： take  s  The words in are sorted according to their initial positions , obtain  "Me1 Myself2 and3 I4" , Then delete the number .

Tips ：

• 2 <= s.length <= 200
• s Only lowercase and uppercase letters 、 Spaces and from 1 To 9 The number of .
• s The number of words in is 1 To 9 individual .
• s Words in are separated by a single space .
• s Does not contain any leading or suffix spaces .

---

Ideas ：

Thinking is more violent haha , Is to define a variable first cur Find the position of the number , And then use begin Variable from cur From back to front until you find Space , Or if it's the first word , Will be out of bounds , So you have to judge . And then begin To cur - 1 At the end of the letter insert the new string Variable tmp in , then Add a space at the end .

Then cycle through it , Add one to the number you want to find each time , Until the number cannot be found ,cur Out of bounds until . Finally, delete the last space at the end .

class Solution {
    
public:
    string sortSentence(string s) {
    
        string tmp;
        char i = '1'; // For viewing s Numeric characters in 
        size_t cur = s.find(i); // use cur Look for numbers 
        while(cur != string::npos)
        {
    
            size_t begin = s.rfind(' ', cur); // use begin Find the beginning of the word 

            // if begin Come to the beginning , Then directly start with cur Part of append that will do 
            if(begin == string::npos)
            {
    
                tmp.append(s, 0, cur);
            }
            else
            {
    
                tmp.append(s, begin + 1, cur - begin - 1);
            }
            i++;
            cur = s.find(i);
            tmp.push_back(' '); // Remember to give each time tmp Insert a space at the end 
        }
        tmp.pop_back(); // Remove the last space 
        return tmp;
    }
};