反转一个字符串中所有单词的位置

题目：给定一个字符串,逐个翻转字符串中的每个单词。
输入:" the sky is b|ue" 输出:" blue is sky the"
示例2
输入:" hello world! 输出:“Word!helo”
解释:输入字符串可以在前面或者后面包含多余的空格,但是反转后的字符不能包括。
示例3
输入:" a good example" 输出:" example good a"
解释:如果两个单词间有多余的空格,将反转后单词间的空格减少到只含一个。

分析：去除空格用双指针的写法，反转每个单词以及反转整个字符串可以达到反转单词位置的效果

#include "_myPrint.cpp"
using namespace std;

class Solution{
public:
    string removeExtraSpaces(string s){
        int slowIndex = 0, fastIndex = 0;
        // begin space
        while (s.size() > 0 && fastIndex < s.size() && s[fastIndex] == ' '){
            fastIndex++;
        }
        // middle space, just save one
        for (; fastIndex < s.size(); fastIndex++){
            if (fastIndex >= 1 && s[fastIndex] == s[fastIndex - 1] && s[fastIndex] == ' '){
                continue;
            }
            s[slowIndex++] = s[fastIndex];
        }
        // end space, may have one or not
        if (slowIndex >= 1 && s[slowIndex - 1] == ' '){
            s.resize(slowIndex - 1);
        }else{
            s.resize(slowIndex);
        }
        return s;
    }
    string reverseEveryWord(string s){
        int slowIndex = 0, fastIndex = 0;
        for (; fastIndex < s.size(); fastIndex++){
            if (s[fastIndex] == ' '){
                reverse(s.begin() + slowIndex, s.begin() + fastIndex);
                slowIndex = fastIndex + 1; // [ )
            }
        }
        reverse(s.begin() + slowIndex, s.end());
        return s;
    }
    string reverseWholeStr(string s){
        reverse(s.begin(), s.end());
        return s;
    }

};

int main(){
    Solution sl;
    string s = "  regdh  fsfs ghdjl   ";
    s = sl.removeExtraSpaces(s);
    s = sl.reverseEveryWord(s);
    s = sl.reverseWholeStr(s);
    cout << s << '$' << endl;

}

标签：slowIndex,begin,string,反转,fastIndex,单词,&&,字符串
来源： https://blog.csdn.net/weixin_43164504/article/details/119909602