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LeetCode:Reverse Words in a String

题目:Reverse Words in a String

Given an input string, reverse the string word by word. For example, Given s = "the sky is blue", return "blue is sky the". 

比较基础的一个题,拿到这个题,我的第一想法是利用vector来存每一个子串,然后在输出,这是一个比较简单的思路,此外,还有第二个思路,就是对所有的字符反转,然后在针对每一个子串反转,但是这个时候要注意它的要求,保证每个子串中只有一个空格。我是按照第一种思路,代码如下:

 1 void reverseWords(string &s) 2 { 3     int i = 0, j = 0; 4     string subStr; 5     vector<string> vecStr; 6     for (j = 0; j != s.length()+1; ++j) { 7         if (s[j] ==  ||j == s.length()) { //Ensure that the final substr can be get 8             subStr = s.substr(i, j - i); 9             if (subStr != "")  //remove the "" from begin and end str10                 vecStr.push_back(subStr);11             i = j + 1;12         }13     }14     15     int vecLen = vecStr.size();16     if (vecLen > 0) {  // deal with the s = ""17         string strResult = "";18         for (i = vecLen - 1; i > 0; i --) {19             strResult += vecStr[i] + " ";20         }21         strResult += vecStr[i];22         s = strResult;23     }24     else25         s = "";26 }

测试情况注意几种:首尾有" "的情况;有多个" "的情况;s = ""的情况;

另外,看到有网友zhangyuehuan的专栏提供了一种更为简洁的思路:

从字符串的最后一个字符遍历,遇到空格就保存子串,然后再对子串反转,和我上面的思路类似,只不过我的遍历方法是正向遍历的,但是其代码简洁,值得学习:

void reverseWords(string & s){    string ss;    int i = s.length()-1;    while(i>=0)    {        while(i>=0&&s[i] ==  ) //处理多个空格的情况        {            i --;        }        if(i<0) break;        if(ss.length()!=0)            ss.push_back( );        string temp ;        for(;i>=0&&s[i]!= ;i--)            temp.push_back(s[i]);        reverse(temp.begin(),temp.end());        ss.append(temp);    }    s=ss;}

 

LeetCode:Reverse Words in a String