2000. reverse word prefix

2000. Reverse word prefix

I'll give you a subscript from 0 Starting string word And a character ch . find ch First occurrence of subscript i , reverse word From the subscript 0 Start 、 Until the subscript i end （ Including subscript i ） That part of the character . If word There are no characters in ch , Then no operation is required .

for example , If word = "abcdefd" And ch = "d" , Then you should reverse From the subscript 0 Start 、 Until the subscript 3 end （ Including subscript 3 ）. The result string will be "dcbaefd" .

return Result string .

Example 1：

 Input ：word = "abcdefd", ch = "d"
 Output ："dcbaefd"
 explain ："d"  The first time it appears in the subscript  3 . 
 Reverse subscript  0  To the subscript  3（ Including subscript  3） This character , The result string is  "dcbaefd" .

Example 2：

 Input ：word = "xyxzxe", ch = "z"
 Output ："zxyxxe"
 explain ："z"  The first and only occurrence is in the subscript  3 .
 Reverse subscript  0  To the subscript  3（ Including subscript  3） This character , The result string is  "zxyxxe" .

Example 3：

 Input ：word = "abcd", ch = "z"
 Output ："abcd"
 explain ："z"  There is no in  word  in .
 No inversion is required , The result string is  "abcd" .

Their thinking :

1. Find the position of the first character in the string
2. Determine whether the character can be found
   • If the character is not found , Just return the original string directly
   • If the character is found , After the character is reversed from the beginning to the position found , Return string

Code :

// Because it's easier , So just use algorithm
class Solution 
{
    
public:
    string reversePrefix(string word, char ch)
    {
    
        // Find the subscript of the first matching character in the string , If you can't find it, go back -1
        auto it = word.find_first_of(ch);	
        // Reverse the character from the start position to the current position .  Because the return cannot be found -1, -1 + 1 = 0; So there's no need to judge 
        reverse(word.begin(), word.begin() + it + 1);
        // Just return the string ;
        return word;
    }
};