Problem

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

Example

"/home/", => "/home" //去掉末尾的slash"/a/./b/../../c/", => "/c" //每個(gè)"/../"對(duì)應(yīng)：刪除一個(gè)上層的segment

Challenge

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".

Note

關(guān)于challenge的兩點(diǎn)：

"/../"，這里討論的有兩種情況，空集和"/../"本身?？占右粋€(gè)if語句返回slash就可以了，"/../"本身要綜合Example的例子，pop出上一層元素。

Multiple slashes，可以用split()函數(shù)去掉所有slash，然后多考慮一個(gè)slash中間為空的case。

關(guān)于switch語句的一個(gè)特點(diǎn)：

我們對(duì)case ""和case "."其實(shí)都是不做操作，然而，兩個(gè)case可以都break，但是不能都不做操作。像這樣：

case "": case ".":

這樣這兩個(gè)case就都會(huì)等價(jià)于下一個(gè)case：case "..". 就會(huì)出錯(cuò)。

Solution

public class Solution {public String simplifyPath(String path) {Stack<String> stack = new Stack<String>();String[] segments = path.split("/");for (String segment: segments) {switch(segment) {case "": break;case ".":case "..": if (!stack.isEmpty()) {stack.pop();}break;default: stack.push(segment);}}StringBuilder sb = new StringBuilder();if (stack.isEmpty()) {//空集的情況return "/";}while (!stack.isEmpty()) {sb.insert(0, "/"+stack.pop());//Don't miss the slash!}return sb.toString();} }

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