C++程序反转给定字符串中的单词

示例：让输入字符串为“我非常喜欢这个程序”。该函数应将字符串更改为“much very program this like i”

例子：

Input: s = “geeks quiz practice code” 
Output: s = “code practice quiz geeks”

Input: s = “getting good at coding needs a lot of practice” 
Output: s = “practice of lot a needs coding at good getting”

算法：

• 最初，将给定字符串的单个单词逐个反转，对于上面的示例，在反转单个单词之后，字符串应该是“i ekil siht margorp yrev hcum”。
• 在上面的示例中，从头到尾反转整个字符串以获得所需的输出“much very program this like i”。

下面是上述方法的实现：

C++

// C++ program to reverse a string
#include 
using namespace std;
  
// Function to reverse words
void reverseWords(string s)
{    
    // Temporary vector to store all words
    vector tmp;
    string str = "";
    for (int i = 0; i < s.length(); i++) 
    {        
        // Check if we encounter space 
        // push word(str) to vector
        // and make str NULL
        if (s[i] == ' ') 
        {
            tmp.push_back(str);
            str = "";
        }
  
        // Else add character to 
        // str to form current word
        else
            str += s[i];
    }
    
    // Last word remaining,add it to vector
    tmp.push_back(str);
  
    // Now print from last to first in vector
    int i;
    for (i = tmp.size() - 1; i > 0; i--)
        cout << tmp[i] << " ";
    // Last word remaining,print it
    cout << tmp[0] << endl;
}
  
// Driver Code
int main()
{
    string s = 
    "i like this program very much";
    reverseWords(s);
    return 0;
}

输出：

much very program this like i

上面的代码不处理字符串以空格开头的情况。以下版本处理了这种特定情况，并且在中间有多个空格的情况下不会对 reverse函数进行不必要的调用。感谢 rka143 提供这个版本。

C++

// C++ program to implement
// the above approach
void reverseWords(char* s)
{
  char* word_begin = NULL;
  
  // temp is for word boundary 
  char* temp = s;
  
  // STEP 1 of the above algorithm 
  while (*temp) 
  {
    /*This condition is to make sure that 
      the string start with valid character 
      (not space) only*/
    if ((word_begin == NULL) && 
        (*temp != ' ')) 
    {
      word_begin = temp;
    }
    if (word_begin && 
       ((*(temp + 1) == ' ') ||
       (*(temp + 1) == ''))) 
    {
      reverse(word_begin, temp);
      word_begin = NULL;
    }
    temp++;
  
  // End of while 
  } 
  
  // STEP 2 of the above algorithm 
  reverse(s, temp - 1);
}
  
// This code is contributed by rutvik_56.

时间复杂度： O(n)

另一种方法：

我们可以通过以相反的方式拆分和保存字符串来完成上述任务。

下面是上述方法的实现：

C++

// C++ program to reverse a string
// s = input()
#include 
using namespace std;
  
// Driver code
int main()
{
    string s[] = {"i", "like", "this", 
                  "program", "very", "much"};
                                          
    string ans = "";
    for (int i = 5; i >= 0; i--) 
    {
        ans += s[i] + " ";
    }
    cout << 
    ("Reversed String:") << endl;
    cout << 
    (ans.substr(0, ans.length() - 1)) << 
     endl;
  
    return 0;
}
// This code is contributed by divyeshrabadiya07.

输出：

Reversed String:
much very program this like i

时间复杂度： O(n)

不使用任何额外空间：
上面的任务也可以通过从中间开始拆分和直接交换字符串来完成。由于涉及直接交换，因此消耗的空间也更少。

下面是上述方法的实现：

C++

// C++ code to reverse a string
#include 
using namespace std;
  
// Reverse the string
string RevString(string s[], int l)
{    
    // Check if number of words is even
    if (l % 2 == 0)
    {        
        // Find the middle word
        int j = l / 2;
          
        // Starting from the middle
        // start swapping words at 
        // jth position and l-1-j position
        while (j <= l - 1)
        {
            string temp;
            temp = s[l - j - 1];
            s[l - j - 1] = s[j];
            s[j] = temp;
            j += 1;
        }
    }
      
    // Check if number of words is odd
    else
    {        
        // Find the middle word
        int j = (l / 2) + 1;
          
        // Starting from the middle start
        // swapping the words at jth 
        // position and l-1-j position
        while (j <= l - 1) 
        {
            string temp;
            temp = s[l - j - 1];
            s[l - j - 1] = s[j];
            s[j] = temp;
            j += 1;
        }
    }
      
    string S = s[0];
      
    // Return the reversed sentence
    for(int i = 1; i < 9; i++)
    {
        S = S + " " + s[i];
    }
    return S;
}
  
// Driver code
int main()
{
    string s = "getting good at coding "
               "needs a lot of practice";
                 
    string words[] = {"getting", "good", "at", 
                      "coding", "needs", "a", 
                      "lot", "of", "practice"};
       
    cout << RevString(words, 9) << endl;
    return 0;
}
  
// This code is contributed by divyesh072019

输出：

practice of lot a needs coding at good getting

另一个直观的常量空间解决方案：遍历字符串并镜像字符串中的每个单词，然后在最后镜像整个字符串。

以下 C++ 代码可以处理多个连续空格。

C++

// C++ program to implement
// the above approach
#include 
#include 
#include 
 using namespace std;
   
string reverse_words(string s)
{
    int left = 0, i = 0, n = s.size();
     
    while (s[i] == ' ')
        i++;
     
    left = i;
     
    while (i < n)
    {
        if (i + 1 == n || 
            s[i] == ' ')
        {
            int j = i - 1;
            if (i + 1 == n)
                j++;
             
            while (left < j)
                swap(s[left++], s[j--]);
             
            left = i + 1;
        }
        if (s[left] == ' ' && i > left)
            left = i;
         
        i++;
    }
    reverse(s.begin(), s.end());
    return s;
}
  
// Driver code
int main()
{ 
    string str = 
    "Be a game changer the world is already full of players"; 
    str = reverse_words(str); 
    cout << str;   
    return 0;
}
// This code is contributed by Gatea David

输出：

players of full already is world the changer game a Be

有关详细信息，请参阅有关给定字符串中的反向单词的完整文章！