lc 简化路径

Posted 2021-02-19 friskypuppy

链接：https://leetcode-cn.com/explore/interview/card/bytedance/242/string/1013/

代码：

#include <stack>
class Solution {
public:
    string simplifyPath(string path) {
        int len = path.size();
        int i = 0;
        stack<string> s;
        while(i < len) {
            while(i < len && path[i] == ‘/‘) {
                i++;
            }
            int flag1 = 0;
            int flag2 = 0;
            int flag3 = 0;
            if((i<len&&i+1<len&&i+2<len) && path[i]==‘.‘ && path[i+1] == ‘.‘ && path[i+2] != ‘/‘) flag1 = 1;
            if((i<len&&i+1<len) && path[i] == ‘.‘ && path[i+1] != ‘/‘ && path[i+1] != ‘.‘) flag2 = 1;
            if(i < len && path[i] != ‘.‘) flag3 = 1;
            if(flag1 || flag2 || flag3) {
                cout << "flag123" << endl;
                // in stack
                string temp = "";
                while(i < len && path[i] != ‘/‘) {
                    temp += path[i];
                    i++;
                }
                s.push(temp);
                continue;
            }
            int flag4 = 0;
            if((i<len&&i+1<len) && path[i] == ‘.‘ && path[i+1] == ‘.‘) flag4 = 1;
            if(flag4) {
                cout << "flag4" << endl;
                // .. if valid, back to previous layer
                if(!s.empty()) {
                    s.pop();
                }
                i += 2;
                continue;
            }
            int flag5 = 0;
            if(i < len && path[i] == ‘.‘) flag5 = 1;
            if(flag5) {
                cout << "flag5" << endl;
                // . stay the same
                i++;
                continue;
            }
        }
        if(s.empty()) return "/";
        stack<string> ans;
        while(!s.empty()) {
            ans.push(s.top());
            s.pop();
        }
        string res = "";
        while(!ans.empty()) {
            res += "/";
            res += ans.top();
            ans.pop();
        }
        return res;
    }
};

View Code

思路：字符串切割，注意"..*" ".*" "../" "./"情况要好好讨论，不能先讨论前缀，再运用栈（主要是".."类似于弹栈操作）输出。