[LeetCode] Reverse Words in a String

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

".

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：用s.substr(pos,num)把s中的子字符串取出来，存放在栈strStack中

然后把每个子字符串从strStack中取出来组合成s；

注意事项：

1.s.subStr(0,0)的结果是“”，即空字符串；

2.测试用例：

（1）字符串首有空格、字符串末尾有空格、字符串中间有连续多个空格

（2）空字符串

（3）只有空格的字符串

class Solution
{
public:
void reverseWords(string &s)
{
if(s.size() == 0)
return;
string::size_type pos=0,num=0,n=0;
stack<string> strStack;

while(n < s.size())
{
while( n < s.size() && s
== ' ')
{
++n;
}
string::size_type num=0;
if( n < s.size() && s
!= ' ')     //找子字符串的初始位置pos，有可能找不到，pos=0
{
pos = n;
++num;
++n;
}
while( n < s.size() && s
!= ' ')  //计算子字符串的长度num，有可能num=0
{
++n;
++num;
}
string subS = s.substr(pos,num); // s.substr(0，0)=""
if(subS != "")
strStack.push(subS);
}

if(strStack.empty()) //有可能原s=“   ”（几个空格），栈中什么都没有，则s就不会从栈中得到但期望的结果是""（空字符串）。
{
s = "";
}
else
{
s = strStack.top();
strStack.pop();
}

while(!strStack.empty())
{
s += ' ';
s += strStack.top();
strStack.pop();
}

}
};