[抄题]:

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

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

[暴力解法]:

时间分析:

空间分析:

[优化后]:

时间分析:

空间分析:

[奇葩输出条件]:

[奇葩corner case]:

空格属于不需要输出的特例,所以先存到hashset中

deque中啥也没有,就输出/

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".

[思维问题]:

[一句话思路]:

用push pop从一端把该存的存入,用for循环从另一端输出

[输入量]:空: 正常情况:特大:特小:程序里处理到的特殊情况:异常情况(不合法不合理的输入):

[画图]:

[一刷]:

  1. hashset<>(),括号中表示要构造的内容,把array函数放进去
Arrays.asList("..",".","")

[二刷]:

  1. deque非空才能POP

[三刷]:

[四刷]:

[五刷]:

[五分钟肉眼debug的结果]:

[总结]:

用push pop从一端把该存的存入,用for循环从另一端输出

[复杂度]:Time complexity: O(n) Space complexity: O(n)

[英文数据结构或算法,为什么不用别的数据结构或算法]:

从左往右走:

"/b/c/" - directory 'b ' - > directory 'c '
"." - current directory
"./" - current directory
"../" - one directory up
e.g
"/" : root directory
"b/c/../" : it will go from c to b
"c/b/./" : it is still in directory b

[算法思想:递归/分治/贪心]:

[关键模板化代码]:

[其他解法]:

[Follow Up]:

[LC给出的题目变变变]:

[代码风格] :

class Solution {
public String simplifyPath(String path) {
//cc //ini: deque, set,put into deque
Deque<String> deque = new LinkedList<>();
Set<String> set = new HashSet<>(Arrays.asList("",".",".."));
for (String w : path.split("/")) {
if (w.equals("..") && !deque.isEmpty()) deque.pop();
else if (!set.contains(w)) deque.push(w);
} //get from deque, res
String res = "";
for (String words : deque)
res = "/" + words + res;
return res.isEmpty() ? "/" : res;
}
}

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