Simplify Path--LeetCode

题目：

Given an absolute path for a file (Unix-style), simplify it.

For example,
path = "/home/", => "/home"
path = "/a/./b/../../c/", => "/c"

click to show corner cases.

Corner Cases:

Did you consider the case where path = "/../"?
In this case, you should return "/".
Another corner case is the path might contain multiple slashes '/' together, such as "/home//foo/".
In this case, you should ignore redundant slashes and return "/home/foo".

思路：使用栈辅助数据结构，来遍历整个字符串，注意在遍历字符串，根据当前字符和栈顶的字符来消除栈空间的内容，最后栈中留下的东西正好是最终的路径，不过觉得使用向量也非常合适

#include <iostream>
#include <string>
#include <vector>
#include <stack>
using namespace std;
/*
给一个Unix下面的路径表示方式 
给出最终的结果 使用栈来消除复杂的表达方式 
*/
string SimplifyPath(string& str)
{
    string result;
    stack<char> st;
    if(str.length() == 0 || str[0] !='/')
        return result;
    int i;
    char tmp;
    st.push(str[0]);
    for(i=1;i<str.length();i++)
    {
        tmp = st.top();
        if(isalpha(str[i])) //是字母 
            st.push(str[i]);
        if(str[i]=='/' && i!=str.length()-1)
        {
            if(tmp != '/' && tmp != '.') // 斜线 
            {
                st.push(str[i]);
            }    
            if(tmp == '.')
            {
                st.pop();
            }
        }
        if(str[i]=='.') //逗点 
        {
            if(tmp == '.')
            {
                st.pop();
                st.pop();
                if(!st.empty())
                    tmp = st.top();
                while(!st.empty() && tmp != '/')
                {
                    st.pop();
                    tmp = st.top();
                }
                if(st.empty())
                    st.push('/');
            }
            else
                st.push(str[i]);
        }
    }
    result.append(st.size(),'c');
    i=st.size()-1;    
    while(!st.empty())
    {
        tmp = st.top();
        st.pop();
        result[i--] = tmp;
    }
    return result;
} 
int main()
{
    string str("/a/./b/../../c/");
    //string str("/../");
    cout<<SimplifyPath(str); 
    return 0;
}