LeetCode 71. Simplify Path（简化路径）

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原题网址：https://leetcode.com/problems/simplify-path/

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

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

click to show corner cases.

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".

方法：通过栈来保存当前路径。

public class Solution {    /*"./""/""/sa///./aa""/sa////aa""/.""a/b/c/d///""/a/./b/../../c/""/home/../../.."*/    public String simplifyPath(String path) {        char[] pa = path.toCharArray();        int pos = 0;        for(int i=0; i<pa.length; i++) {            if (pa[i] == '/') {                if (pos>0 && pa[pos-1] != '/') pa[pos++] = pa[i];            } else if (pa[i] == '.') {                if ((((pos==0) || (pos>0 && pa[pos-1] == '/')) && i+2<pa.length && pa[i+1]=='.' && pa[i+2]=='/')                || (((pos==0) || (pos>0 && pa[pos-1] == '/')) && i+2==pa.length && pa[i+1]=='.')) {                    //upper level                    pos--;                    while (pos>0 && pa[pos-1] != '/') pos --;                    if (pos<0) pos = 0;                    i ++;                } else if (((pos==0) || (pos>0 && pa[pos-1] == '/')) && (i+1==pa.length || (i+1<pa.length && pa[i+1] == '/'))) {                    // same level                } else {                    // normal dir                    pa[pos++] = pa[i];                }            } else {                pa[pos++] = pa[i];            }        }        if (pos>0 && pa[pos-1] == '/') pos --;        char[] simplified = new char[pos+1];        simplified[0] = '/';        for(int i=0; i<pos; i++) simplified[i+1] = pa[i];        return new String(simplified);    }}

词法和语法分开可以简化：

public class Solution {    public String simplifyPath(String path) {        String[] paths = path.split("/");        int len = 0;        for(int i=0; i<paths.length; i++) {            if ("..".equals(paths[i])) {                len = Math.max(--len, 0);            } else if (!"".equals(paths[i]) && !".".equals(paths[i])) {                paths[len++] = paths[i];            }        }        StringBuilder sb = new StringBuilder("/");        for(int i=0; i<len; i++) {            sb.append(paths[i]);            if (i<len-1) sb.append("/");        }        return sb.toString();    }}

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