[算法分析与设计] leetcode 每周一题: 071. Simplify Path

题目链接: 

071. Simplify Path

题目大意:

给定一个 Unix 风格的 path, 给出一个简化的版本 ;

题中还注明了 对顶层目录 的假设, 见下方 例3 ;

例如: 给定: "/home/", 则输出应为: "/home" ;

例如: 给定: "/a/./b/../../c/", 则输出应为: "/c" ;

例如: 给定: "/../", 则输出应为: "/" ;

解题过程:

(1) 题意很简单, 由于关键是对 corner case 的处理, 所以我大致上就直接按工程的做法做了 ;

(2) 就用最老实的做法, 逐项处理 ; 针对 "当前目录" 符号 "./", 直接忽略 ; 针对 "父目录" 符号 "../" 考虑利用栈来处理 (中间还要假设顶层目录是 "/") ;

代码如下:

class Solution {  public:    using Path = string;    Path withoutRedundantSlash( const Path & s ) {        Path ret;        ret.push_back( s.front() );        for ( size_t i = 1; i < s.size(); i++ ) {            if ( s[i] == '/' && s[i - 1] == '/' ) {                continue;            }            ret.push_back( s[i] );        }        return ret;    }    Path withoutTailSlash( const Path & s ) {        auto ret = s;        while ( ret.size() > 1 && ret.back() == '/' ) {            ret.pop_back();        }        return ret;    }    Path withoutRedundantDot( const Path & s ) {        vector< Path > segments;        size_t head = 0;        for ( size_t i = head; i < s.size(); i++ ) {            if ( s[i] == '/' ) {                segments.push_back( s.substr( head, i - head ) );                head = i + 1;            }        }        if ( head < s.size() ) {            segments.push_back( s.substr( head ) );        }        vector< Path > dirs;        for ( size_t i = 0; i < segments.size(); i++ ) {            auto d = segments[i];            if ( d == ".." ) {                if ( dirs.size() <= 1 ) {                    // FIXME: ignore heading `./` case                    continue;                }                dirs.pop_back();            } else if ( d == "." ) {                continue;            } else {                dirs.push_back( d + "/" );            }        }        Path ret;        for ( size_t i = 0; i < dirs.size(); i++ ) {            auto d = dirs[i];            ret += d;        }        return ret;    }    // FIXME: assume filepath    Path simplifyPath( Path path ) {        Path ret = path;        // cerr << ret << endl;        ret = withoutRedundantSlash( ret );        // cerr << ret << endl;        ret = withoutRedundantDot( ret );        // cerr << ret << endl;        ret = withoutTailSlash( ret );        return ret;    }};

Runtime: 6 ms