[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。可是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(假设有多个空格)。

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能够AC。
大致分为两步：一个是常规的翻转字符串中的单词；还有一个就是想方法去掉串中的多余的单词；这两步骤的顺序能够颠倒。

以下给出两份代码。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]!=''; ++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] = '';
        else
              work[j] = '';
          //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 )