[LeetCode] 71. Simplify Path

[LeetCode] 71. Simplify Path

---

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”.

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分析文件路徑，其實通過’/’就是分割字符串，’/’之間的字符串分三種情況：

• s == “..”， 保存文件路徑的棧的棧頂出棧。
• s == “.”， 什麼都不做。
• s == 其他， 壓棧。

---

class Solution {
public:
    string simplifyPath(string path) {
        vector<string> p;
        int len = path.length();
        string s = "";
        for (int i=0; i<len; ++i) {
            if (path[i] == '/') {
                if (s == "..") {
                    if (!p.empty()) {
                        p.pop_back();
                    }
                } else if (s != "." && s != "") {
                    p.push_back(s);
                }
                s = "";
            } else {
                s += path[i];
            }
        }
        if (s == "..") {
            if (!p.empty()) {
                p.pop_back();
            }
        } else if (s != "." && s != "") {
            p.push_back(s);
        }
        string res = "";
        int psize = p.size();
        for (int i=0; i<psize; ++i) {
            res += "/" + p[i];
        }
        return res == "" ? "/" : res;
    }
};