leetcode做题总结，题目Simplify Path 71

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题目是Simplify Path

这道题思路很简单，就是按照“/”分开然后压栈，遇到特定字符进行处理即可

public class Solution {    public String simplifyPath(String path) {        if(path.length()==0)            return null;        if(path.length()==1)            return path;        String[] pa = path.split("/");        Stack<String> s = new Stack<String>();        for(int i=0;i<pa.length;i++){            //cannot use pa[i]!=null here, because null is different with "";            if(!pa[i].equals("")){                if(s.size()==0){                    if(pa[i].equals(".")||pa[i].equals(".."));                    else                        s.push(pa[i]);                }else{                    if(pa[i].equals("."))                        continue;                    if(pa[i].equals("..")){                        s.pop();                        continue;                    }                    s.push(pa[i]);                }                                                }                    }        String res="";        int len=s.size();        //when using collection's length in for, make sure the length will NOT change!        for(int i=0;i<len;i++){            res = "/"+s.pop() + res;        }        if(res.length()==0)            return "/";        return res;    }}

Update 2015、08、31：思路一样，下面是九章的代码，感觉逻辑能简单些

public class Solution {    /**     * @param path the original path     * @return the simplified path     */    public String simplifyPath(String path) {        // Write your code here        String result = "/";        String[] stubs = path.split("/+");        ArrayList<String> paths = new ArrayList<String>();        for (String s : stubs){            if(s.equals("..")){                if(paths.size() > 0){                    paths.remove(paths.size() - 1);                }            }            else if (!s.equals(".") && !s.equals("")){                paths.add(s);            }        }        for (String s : paths){            result += s + "/";        }        if (result.length() > 1)            result = result.substring(0, result.length() - 1);        return result;    }}

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