反转字符串中单词的顺序 [英] Reverse the ordering of words in a string

问题描述

我有这个字符串s1 =我的名字是XYZ，我想扭转单词的顺序，以便 s1 =ZYX是我的名字。

我可以使用额外的数组来做。我想很难，但是有可能做到这一点（不使用额外的数据结构），时间复杂度是O（n）？

解决方案

反转字符串，然后在第二遍中，反向每个单词...

在c＃中，完全就地没有其他数组：

  static char [] ReverseAllWords（char [] in_text）
 {
 int lindex = 0; 
 int rindex = in_text.Length  -  1; 
 if（rindex> 1）
 {
 // reverse complete phrase 
 in_text = ReverseString（in_text，0，rindex）; 
 
 //反转每个单词的结果反转短语
 for（rindex = 0; rindex <= in_text.Length; rindex ++）
 {
 if（rindex = = in_text.Length || in_text [rindex] ==''）
 {
 in_text = ReverseString（in_text，lindex，rindex  -  1）; 
 lindex = rindex + 1; 
} 
} 
} 
 return in_text; 
} 
 
 static char [] ReverseString（char [] intext，int lindex，int rindex）
 {
 char tempc; 
 while（lindex< rindex）
 {
 tempc = intext [lindex]; 
 intext [lindex ++] = intext [rindex]; 
 intext [rindex--] = tempc; 
} 
 return intext; 
} 
  

I have this string s1 = "My name is X Y Z" and I want to reverse the order of the words so that s1 = "Z Y X is name My".

I can do it using an additional array. I thought hard but is it possible to do it inplace (without using additional data structures) and with the time complexity being O(n)?

解决方案

reverse the string and then, in a second pass, reverse each word...

in c#, completely in-place without additional arrays:

static char[] ReverseAllWords(char[] in_text)
{
    int lindex = 0;
    int rindex = in_text.Length - 1;
    if (rindex > 1)
    {
        //reverse complete phrase
        in_text = ReverseString(in_text, 0, rindex);

        //reverse each word in resultant reversed phrase
        for (rindex = 0; rindex <= in_text.Length; rindex++)
        {
            if (rindex == in_text.Length || in_text[rindex] == ' ')
            {
                in_text = ReverseString(in_text, lindex, rindex - 1);
                lindex = rindex + 1;
            }
        }
    }
    return in_text;
}

static char[] ReverseString(char[] intext, int lindex, int rindex)
{
    char tempc;
    while (lindex < rindex)
    {
        tempc = intext[lindex];
        intext[lindex++] = intext[rindex];
        intext[rindex--] = tempc;
    }
    return intext;
}

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