[leetcode] Reverse Words in a String [1]

一、题目：

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

For example,

Given s = "

the sky is blue

",

return "

blue is sky the

".

click to show clarification.

Clarification:

What constitutes a word?

A sequence of non-space characters constitutes a word.
Could the input string contain leading or trailing spaces?

Yes. However, your reversed string should not contain leading or trailing spaces.
How about multiple spaces between two words?

Reduce them to a single space in the reversed string.

原题地址

二、解题思想

翻转字符串中的单词顺序，这是个老题目了，但是leetcode上面的要求更为严格，如：

要求把开头和结尾的空格删除掉；

缩减单词间的空格数为1(如果有多个空格)；

单词若全是空格，则返回一个空字符串("").

此题思想不难，主要是注意上面三个要求和一些细节就可以AC。

大致分为两步：一个是常规的翻转字符串中的单词；另一个就是想方法去掉串中的多余的单词；这两步骤的顺序可以颠倒。

下面给出两份代码，第一个代码是先去掉多余的空格，然后在翻转；第二个代码先翻转，在去掉多余的空格。就效率上来说应该是第一个代码的效率更高。

三、代码实现

代码一：

class Solution {
public:
    void reverseWords(string &s) {
        if(s.size()<=0) return ;
        char *work = new char[s.size()+1];
        //reduce blank
        int j=0;
        for(int i=0; s[i]!='\0'; ++i){
            if(i>0 && s[i] == ' ' && s[i-1]!= ' ')
                work[j++] = s[i];
              else if(s[i] != ' ')
                  work[j++] = s[i];
          }
          if(j>0 && work[j-1]==' ')
              work[--j] = '\0';
        else
              work[j] = '\0';
          //reverse all string
          reverse(work, 0, j-1);
          int p= 0, i=0;
          //reverse each word
          while(i<j){
               while(p<j && work[p]!=' ') p++;
               reverse(work, i, p-1);
               i = p+1;
               p = i;               
          }
          string temp(work);
          s = temp;
    }
    
    void reverse(char *s, int beg, int end){
        while(beg < end){
            char temp = s[beg];
            s[beg++] = s[end];
            s[end--] = temp;
        }
    }
};

代码二：

class Solution {
public:
    void reverseWords(string &s) {
        int n = s.size();
        if(n<=0) return;
        //if(n==1)
        //reverse the whole string
        reverse(s, 0, n-1);
        //reverse each word
        int begin=0, end = 0;
        while(begin<n){
            while( begin< n && s[begin] == ' ') ++begin;
            end = begin;
            while( end<n && s[end] != ' ') ++end;
            reverse(s, begin, end-1);
            begin = end;
        }
        //reduce blank
        begin = 0;
        while(begin<n && s[begin] ==' ') ++begin;
        if(begin == n) {s = s.substr(0,0);return;}
        
        end = n-1;
        while(end>=0 && s[end] == ' ') --end;
        
        int start = 0;
        char pre;
        for(; begin<=end; ++begin){
            if(s[begin] != ' '){
                s[start++] = s[begin];
                pre = s[begin];
            }else{
                if(pre != ' '){
                    s[start++] = ' ';
                    pre = ' ';
                }
            }
        }
        if(start != n) s = s.substr(0, start);
    }
    
    void reverse(string &s, int begin, int end){
        char temp;
        while(begin<end){
            temp = s[begin];
            s[begin++] = s[end];
            s[end--] = temp;
        }
    }
};

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（转载文章请注明出处: http://blog.csdn.net/swagle/article/details/28236933 )