Reverse Words in a String(Need edition)

題目

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

For example,
Given s = "the sky is blue",
return "blue is sky the".

分析

字符串處理型。一種方法是從string中分析出每個單詞，需要考慮各種空格問題；另外一種是定義輸入流，把string作爲輸入流，每次讀取一個string，更加簡潔。

時間複雜度

第一種爲O(s.size())；

輸入

空字符串需要特殊處理

""
" " / " "
"a" / "the sky is blue"
" a" /" the sky is blue"
" a "

結論

CODE1 : 48ms

CODE2 : 64ms

CODE3 : _

CODE2僅僅是把CODE1中的插入單詞作爲單獨的函數，函數調用使佔用時間增加?

CODE

CODE1

class Solution {
public:
    void reverseWords(string &s) {
        if(s.size() == 0) return;
        string word, str;
        bool flag = false;
        for (auto c : s) {
            if (!isspace(c)) {
                if (flag == false) {
                    flag = true;
                }
                word += c;
            } else {
                if(flag == true) {
                    if(str.empty()) {
                         str = word;
                    } else {
                        str = word + " " + str;
                    }
                    
                    word = "";
                    flag = false;
                } else {
                    continue;
                }
            }
        }
        if (!isspace(s[s.size()-1])) {
            if(str.empty()) {
                 str = word;
            } else {
                str = word + " " + str;
            }
        }
        s = str;
    }
};

CODE2

class Solution {
public:
    void reverseWords(string &s) {
        if(s.size() == 0) return;
        string word, str;
        // flag: This poit is space?
        // at the beginning, there is no word
        bool flag = false;  
        for (auto c : s) {
            if (!isspace(c)) {
                if (flag == false) {
                    flag = true;
                }
                word += c;
            } else {
                if(flag == true) {
                    insertToString(str, word);
                    flag = false;
                } else {
                    continue;
                }
            }
        }
        if (!isspace(s[s.size()-1])) {
            insertToString(str, word);
        }
        s = str;
    }
    
    void insertToString(string &str, string &word) {
        if(str.empty()) {
            str = word;
        } else {
            str = word + " " + str;
        }
        word = "";
    }
};

CODE3