LeetCode 题解（16）: Simplify Path

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题目：

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

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

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".

题解：

1. 将输入字符串以'/'为分隔符分割成子串保存在字符串数组中

2. 遍历1生成的字符串数组，每当碰到非"."和".."的字符串，就将该字符串前加"/"，并压入到一个字符串堆栈中（用vector实现，方便最后拼接路径）

    每当碰到"."时，直接continue

    每当碰到".."时，堆栈弹出一个元素

3. 遍历完成后，将堆栈中剩余的字符串按先进先出的顺序（用vector可以两端访问元素）拼接成最后的路径

     若堆栈为空，则返回"/"。

class Solution {
public:string simplifyPath(string path) {if(!path.length())return path;vector<string> split;for(int i = 0; i < path.length();){if(path[i] == '/'){i++;continue;}else {string temp = "";while(path[i] != '/' && i < path.length()){temp.insert(temp.length(), 1, path[i++]);}split.push_back(temp);}}vector<string> combine;for(int i = 0; i < split.size(); i++){if(split[i] != "." && split[i] != ".."){string path1 = "";path1.insert(path1.length(), 1, '/');path1.insert(path1.length(), split[i]);combine.push_back(path1);}else if(split[i] == ".." && combine.size()){combine.pop_back();}else{continue;}}string final = "";if(!combine.size()){final.insert(final.length(), 1, '/');}for(int i = 0; i < combine.size(); i++){final.insert(final.length(), combine[i]);}return final;}
};

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