题意：给定一个字符串我要得到简化后的路径。 '…‘代表上个路径，’.'代表当前路径
Input: path = “/home/user/Documents/…/Pictures”
Output: “/home/user/Pictures”
https://leetcode.com/problems/simplify-path/description/
解析：首先这道题肯定是从头到尾遍历来做，难点在于我应该用什么判断语句。
如果结尾能够+一个‘/’判断会少很多。用一个cur变量来保存当前可以进入栈的元素，用一个栈来存放当前所有的路径，最后重组即可。
class Solution {
public:string simplifyPath(string path) {vector<string> st;string cur;string ret;if(path.back() != '/') {path += '/';}for(int i = 0; i < path.size(); i++) {if(path[i] != '/') {cur += path[i];continue;} else {if (cur == "..") {if(st.size()) {st.pop_back();}} else if (cur != "." && cur.size()) {st.push_back(cur);}cur.clear();}}if(!st.size()) return "/";for(auto&p : st) {ret += '/';ret += p;}return ret;}
};
Leetcode 71(错误答案)
这种情形没有考虑到/…hidden/的情形
class Solution {
public:string simplifyPath(string path) {vector<string> st;string ret;for(int i = 0; i < path.size(); i++) {if(isalpha(path[i])) {int j = i;while(isalpha(path[j])) j++;string p = path.substr(i, j-i);st.push_back(p);i = j - 1;}if(path[i] == '.') {int j = i;while((path[j] == '.')) j++;// one dotif(j-i == 1)continue;if(j-i == 2) {if(st.size()) {st.pop_back();}}if(j-i > 2) {string p = path.substr(i, j-i);st.push_back(p);}i = j - 1;}if(path[i] == '/') continue;}if (!st.size()) {return "/";}for(auto&p : st) {ret += '/';ret += p;}return ret;}
};