题目例如以下:
Given an input string, reverse the string word by word.
For example,
Given s = "the sky is blue
",
return "blue is sky the
".
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.
题目要求不仅须要把字符串中的单词进行旋转,并且測试用例中还包括头尾含有空格、单词间空格数大于1个之类的字符串。所以要处理好空格,处理完空格的字符串肯定是小于或等于原字符串长度的,考虑到通过移动字符的方式来减小空格耗时长,我的思路是首先计算出去多余的空格后字符串应有的长度,然后又一次定义一个暂时字符串,开辟终于应有长度的空间。然后将原字符串逆序copy到新定义的字符串变量中,保证头尾多余空格已去除,保证单词间仅保留一个空格。然后将每一个单词再进行一次反转,便实现了题目所要求的功能。AC代码例如以下:
class Solution { public: void reverseWords(string &s) { if(s.empty()) return; int count = 0; int index = 0, indexTemp = 0, begin = 0, end = s.length()-1; while(s[begin] == ' ' && begin<s.length()-1) begin++; while(s[end] == ' ' && end>0) end--; if(end == 0 && s[end] == ' ') { s = ""; return; } else if(end == 0) { s = s.substr(0,1); return; } index = begin; while(index <= end) { if(s[index] == ' ') { count++; while(s[index] == ' ') index++; } count++; index++; } string temp(count,'