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【LeetCode】

LeetCode中关于栈的题目其实不算很多，用到的套路也比较单一。

71 Simplify Path。

题目描述：

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

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

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".

这题其实有很大的实际应用意义，主要是对Linux/Unix系统路径的表达。

用ArrayDeque作栈的解法：

class Solution {
    public String simplifyPath(String path) {
        Deque<String> stack = new ArrayDeque<>();
        Set<String> skip = new HashSet<>(Arrays.asList("..", ".", ""));
        for(String dir : path.split("/")){
            if(dir.equals("..") && !stack.isEmpty())
                stack.pop();
            else if(!skip.contains(dir))
                stack.push(dir);
        }
        String res = "";
        for(String str : stack){
            res = "/" + str + res;
        }
        return res.isEmpty() ? "/" : res;
    }
}

其实上面这种解法很慢，相当的慢，在答案的排名中发现了别人的这种的解法，相当于手动模拟栈吧。

class Solution {
    public String simplifyPath(String path) {
        int l = path.length(), i = 0, k = 0;
        char[] C = new char[l];
        char ch;
        for (; i < l; i++) {
            ch = path.charAt(i);
            if (ch != '/') {
                C[k++] = ch;
                continue;
            }
            while (i < l - 1 && path.charAt(i + 1) == '/')
                i++;
            if (i == l - 1)
                break;
            if ((i < l - 3 && path.charAt(i + 1) == '.' && path.charAt(i + 2) == '.' && path.charAt(i + 3) == '/')
                    || (i == l - 3 && path.charAt(i + 1) == '.' && path.charAt(i + 2) == '.')) {
                while (k > 0 && C[--k] != '/')
                    ;
                i += 2;
            } else if ((i < l - 2 && path.charAt(i + 1) == '.' && path.charAt(i + 2) == '/')
                    || (i == l - 2 && path.charAt(i + 1) == '.')) {
                i++;
            } else
                C[k++] = ch;
        }
        return (k == 0) ? "/" : String.valueOf(C, 0, k);
    }
}