Sword finger offer 05. replace spaces

Please implement a function , Put the string  s  Replace each space in with "%20".

  Method 1 ： Apply for extra space

class Solution {
public:
    // thought ： Create a new string as the returned result , Traversal string , If you encounter ' ', Just add “%20”, If the ' ', Append the traversed characters 
    string replaceSpace(string s) {
        string res = "";
        for(auto ch : s){
            if(ch == ' '){
                res += "%20";
            }
            else{
                res += ch;
            }
        }
        return res;
    }
};
// Time complexity ：O（N） Traversal string .
// Spatial complexity ：O（N） A new string Extra space for type .

Method 2 ： Modify in place

class Solution {
public:
//1.  initialization ： The number of spaces  count , character string  s  The length of  len ;
//2.  Count the number of spaces ： Traverse  s , In case of space  count++ ;
//3.  modify  s  length ： After adding  "%20"  The length of the string after should be  len + 2 * count ;
//4.  Traversal modification in reverse order ：i  Point to the element at the end of the original string , j  Point to the element at the end of the new string ; When  i = j  Jump out when （ Represents that there is no space on the left , There is no need to continue traversing ）;
// When  s[i]  When not a space ： perform  s[j] = s[i] ;
// When  s[i]  When is a space ： Close the string  [j-2, j]  The element of is changed to  "%20" ; Because of the modification  3  Elements , Therefore need  j -= 2 ;
//5.  Return value ： Modified string  s ;

    string replaceSpace(string s) {
        int count = 0; // Used to count the number of spaces 
        int len = s.size();
        // Count the number of spaces in the string 
        for(char c : s) {
            if(c == ' ') {
                count++;
            }
        }
        // Modify the length of the original string 
        s.resize(len + 2 * count);
        // Traversal modification in reverse order 
        // use  i  To traverse the original string in reverse order , use  j  To traverse the expanded string in reverse order 
        for(int i = len - 1, j = s.size() - 1; i < j; i--,j--) {
            if(s[i] != ' '){
                s[j] = s[i];
            }
            else{
                s[j - 2] = '%';
                s[j - 1] = '2';
                s[j] = '0';
                j -= 2;
            }
        }
        return s;
    }
    // Time complexity  O(N) ：  Traversal Statistics 、 Traversal and modification all use  O(N)  Time .
    // Spatial complexity  O(1) ：  Because it's in situ expansion  s  length , Therefore use  O(1)  Extra space .
};