71 LeetCode question: simplified path (Average)

71 LeetCode question: simplified path (Average)

• Title: Unix style to give an absolute file path, you need to simplify it. Or in other words, to convert it to the path specification. In Unix-style file system, a dot represents the current directory itself (.); In addition, two dots (...) indicates a directory to the (parent directory,); both of which may be composed of a relatively complex path section. For more information, see: Absolute path vs Linux / Unix is ​​a relative path. Please note that the specification must always return to the path with a slash / at the beginning, and must have only one directory name between two slashes /. Finally, a directory name (if there is) not end with a /. In addition, the specification represents the shortest path must be an absolute path string.
  
• A thought: think of the significant figures of the 65th title issues, in order to practice exercises that questions of method, I chose the way the state set. I think that even the name of the directory can also contain points (black question mark face ???), when this point can leave, when not to have studied for a long time. . . . Finally, when I am almost exhausted before. . . .

class Solution {
    public int make(char c){
        switch(c){
            case '.' :return 0;
            case '/' :return 1;
            default:
                return 2;
        }
    }

    public String simplifyPath(String path) {
        String res = "";
        char[] ans = new char[path.length()];
        ans[0]='/';
        int state = 0;
        int j = 1;
        int[][] transfer = new int[][]{{ 1, 0, 2},
                                       { 3, 0, 2},
                                       { 1, 0, 2},
                                       { 2, 0, 2}};
        char[] ss = path.toCharArray();
        for(int i=1;i<ss.length;i++){
            int id = make(ss[i]);
            int stateNew=transfer[state][id];
            if(state==0){
                if(id==0){
                    ans[j++]='.';
                }else if(id==2){
                    ans[j++]=ss[i];
                }
            }else if(state==1){
                if(id==0){
                    ans[j++]='.';
                }else if(id==1){
                    if(ans[j-2]=='/'){
                        j=j-1;
                    }else if(i!=ss.length-1){
                        ans[j++]='/';
                    }
                }else if(id==2){
                    ans[j++]=ss[i];
                }
            }else if(state==2){
                if(id==0){
                    ans[j++]='.';
                }else if(id==1 && i!=ss.length-1){
                    ans[j++]='/';
                }else if(id==2){
                    ans[j++]=ss[i];
                }

            }else if(state==3){
                if(id==0){
                    ans[j++]='.';
                }else if(id==1){
                    if(j>4){
                        j=j-4;
                        while(j>0 && ans[j] != '/'){
                            j--;
                        }
                        j++;
                    }else{
                        j=1;
                    } 
                    
                }else if(id==2){
                    ans[j++]=ss[i];
                }
            }

            state=stateNew;
        }
        if(state==3){
            if(j>4){
                j=j-4;
                while(j>1 && ans[j] != '/'){
                    j--;
                }
            }else{
                j=1;
            }
            
        }else if(state==1){
            j=j-1;
        }
        if(j>1 && ans[j-1]=='/') j--;
        for(int k=0;k<j;k++){
            res+=ans[k];
        }
        return res;
    }
}

• Thinking two: methods are the solution to a problem in the stack to, and victims, I will not. Next time I encountered a similar problem with a stack method it again. . . .

public String simplifyPath(String path) {
        String[] s = path.split("/");
        Stack<String> stack = new Stack<>();
        
        for (int i = 0; i < s.length; i++) {
            if (!stack.isEmpty() && s[i].equals(".."))
                stack.pop();
            else if (!s[i].equals("") && !s[i].equals(".") && !s[i].equals(".."))
                stack.push(s[i]);
        }
        if (stack.isEmpty())
            return "/";

        StringBuffer res = new StringBuffer();
        for (int i = 0; i < stack.size(); i++) {
            res.append("/" + stack.get(i));
        }
        return res.toString();
    }

作者：StackOverflow-
链接：https://leetcode-cn.com/problems/simplify-path/solution/java-yi-dong-yi-jie-xiao-lu-gao-by-spirit-9-18/
来源：力扣（LeetCode）
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