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leetcode之倒转一句话单词
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.
思路：
这题的思路不用多说，完全是烂大街的题，不就是先以每个单词为单位，进行翻转，再以整句话为单位进行翻转吗？呵呵，太简单了，以为简单你就能做对那可就大错特错了，题干里的的说明还要求去除首尾空格，并压缩多个空格为一个空格，所以会有N多边界条件要考虑，我竟然编译了N次才通过leetcode OJ，作为一个正式工作一年多的程序员，实在是对不起国家，边界条件的确是考察一个人思维严谨性甚至记忆力的试金石！
代码：
class Solution {
public:
Solution(){}
void reverse(std::string &s, int f, int e)
{
while (e>f)
{
char tmp = s.at(f);
s.at(f) = s.at(e);
s.at(e) = tmp;
f++;
e--;
}
}
void squeeze(std::string &s)
{
int i = 0;
std::string re = "";
char prev = NULL;
bool flag = false;
while (i < s.size())
{
if (s.at(i)!=' ')
{
if (flag&&prev == ' ')
{
re.push_back(' ');
}
re.push_back(s.at(i));
}
if (s.at(i) == ' '&&prev !=NULL &&prev!= ' ')
{
flag = true;
}
prev = s.at(i);
i++;
}
s = re;
}
void reverseWords(std::string &s) {
int i = 0;
int f=0, e=0;
char prev = ' ';
while (i<s.size())
{
if (prev == ' '&&s.at(i) != ' ')
{
f = i;
}
if ((s.at(i) == ' '&&prev != ' ') || (i == s.size() - 1 && s.at(i) != ' '))
{
if (i == s.size() - 1 && s.at(i) != ' ')
{
e = i;
}
else
{
e = i - 1;
}
if (f<e)
{
reverse(s, f, e);
}
}
prev = s.at(i);
i++;
}
reverse(s, 0, s.size() - 1);
squeeze(s);
}
};
leetcode之倒转一句话单词
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