标签：leetcode   算法   

问题描述：

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.

• 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".

基本思想：

此题可以用一个堆栈存储每层文件夹或文件，如果两个/之间的内容是".."就弹出栈顶元素，如果遇到“.”或“”就不变。基本思想就是这样。但要考虑一些corner case，如“/../”,"///"等

代码：

public String simplifyPath(String path) {  //Java
        // return "/";
        path = path.trim();
        if(path.equals("/"))
            return "/";
        
        List<String> stack = new LinkedList<String>();
        
        int top = -1;
        String tmp = path;
        while(!tmp.equals("")){
            if(tmp.startsWith("/"))
                 tmp = tmp.substring(1);
            if(tmp.equals(""))
                break;
            
            if(tmp.contains("/")){
                int pos = tmp.indexOf("/");
                String file = tmp.substring(0,pos);
                tmp = tmp.substring(pos);
                if(file.equals(".")|| file.equals(""))
                    continue;
                if(file.equals("..")){
                    if(top>-1)
                        top--;
                    continue;   
                }
                stack.add(++top,file);
                
            }
            else {
                String file = tmp;
                tmp = "";
                if(file.equals("."))
                    continue;
                if(file.equals("..")){
                    if(top>-1)
                        top--;
                    continue;   
                }
                stack.add(++top,file);
                
            }
        }
        //generate path;
        String rpath = "/";
        for(int i = 0; i <=top ; i++)
            rpath += stack.get(i)+"/";
        if(top >=0)
            rpath = rpath.substring(0,rpath.length()-1);
        return rpath;
        
    }

[leetcode]Simplify Path

标签：leetcode   算法   

原文地址：http://blog.csdn.net/chenlei0630/article/details/41910703