[LeetCode] 151Reverse Words in a String C语言版

Given an input string, reverse the string word by word.

Example:

Input: “the sky is blue”,
Output: “blue is sky the”.
Note:

A word is defined as a sequence of non-space characters.
Input string may contain leading or trailing spaces. However, your reversed string should not contain leading or trailing spaces.
You need to reduce multiple spaces between two words to a single space in the reversed string.
Follow up: For C programmers, try to solve it in-place in O(1) space.

分析：
按题目要求，在O(1)的空间复杂度内，将一个字符串中的单词位置进行首尾交换。
1）字符串前后可能有空格，输出的字符串前后不能有空格。
2）两个词之间可能有多个空格，输出时要去除多余的空格。

由于每个单词的长度不同，而且题目要求空间复杂度为O(1)，所以不能直接交换首尾的两个单词。但可以采用如下的方法：
1）将这个字符串看作一个整体，首尾调换“123 456“ —> “654 321”
2)将这个字符串中的每一个单词，再执行首尾调换，”654 321” –>”456 123”

题目的难点在于字符串前后以及单词之间可能有多余的空格，所以我们首先去除这些空格，方便后面的反转。

enum{
    STATE_BEGIN,
    STATE_SPACE,
    STATE_CAPITAL
};
void  remove_duplicate_space(char *s)
{
    if(s==NULL || *s=='\0')
        return;
    int length = strlen(s);
    int index = 0;
    int state = STATE_BEGIN;
    for(int i=0; i<length; i++){
        if(s[i] == ' '){
            if(state == STATE_CAPITAL){
                state = STATE_SPACE;
            }
        }else{
            if(state == STATE_SPACE){
                s[index++] = ' ';
            }
            s[index++] = s[i];
            state = STATE_CAPITAL;
        }
    }
    s[index] = '\0';

}

然后执行反转

void _reverse(char *pStart, char *pEnd)
{
    if(pStart >= pEnd)
        return;
    while((*pEnd == ' ') && (pEnd >=pStart))
        pEnd--;
    while(pStart < pEnd){
        char c = *pStart;
        *pStart = *pEnd;
        *pEnd = c;
        pStart++;
        pEnd--;
    }
}
void reverseWords(char *s) {
    remove_duplicate_space(s);
   int length = strlen(s);
   if(length <= 1)
       return;
   char *pStart = s;
   char *pEnd = s+length -1;
    _reverse(pStart, pEnd);
    while(pStart while(*pStart == ' ')pStart++;
        char *p=pStart;
        while(*p!= ' ' && *p!='\0') p++;
        _reverse(pStart, p-1);
        pStart = p++;
    }
}