[LeetCode] Simplify Path 简化路径

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.

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"

.

这道题让简化给定的路径，光根据题目中给的那一个例子还真不太好总结出规律，应该再加上两个例子 path =

"/a/./b/../c/"

, =>

"/a/c"和path = 
"/a/./b/c/"
, => [code]"/a/b/c"，

这样我们就可以知道中间是"."的情况直接去掉，是".."时删掉它上面挨着的一个路径，而下面的边界条件给的一些情况中可以得知，如果是空的话返回"/"，如果有多个"/"只保留一个。那么我们可以把路径看做是由一个或多个"/"分割开的众多子字符串，把它们分别提取出来一一处理即可，代码如下：[/code]

解法一

class Solution {
public:
string simplifyPath(string path) {
vector<string> v;
int i = 0;
while (i < path.size()) {
while (path[i] == '/' && i < path.size()) ++i;
if (i == path.size()) break;
int start = i;
while (path[i] != '/' && i < path.size()) ++i;
int end = i - 1;
string s = path.substr(start, end - start + 1);
if (s == "..") {
if (!v.empty()) v.pop_back();
} else if (s != ".") {
v.push_back(s);
}
}
if (v.empty()) return "/";
string res;
for (int i = 0; i < v.size(); ++i) {
res += '/' + v[i];
}
return res;
}
};

还有一种解法是利用了C语言中的函数strtok来分隔字符串，但是需要把string和char*类型相互转换，转换方法请猛戳这里。除了这块不同，其余的思想和上面那种解法相同，代码如下：

解法二

class Solution {
public:
string simplifyPath(string path) {
vector<string> v;
char *cstr = new char[path.length() + 1];
strcpy(cstr, path.c_str());
char *pch = strtok(cstr, "/");
while (pch != NULL) {
string p = string(pch);
if (p == "..") {
if (!v.empty()) v.pop_back();
} else if (p != ".") {
v.push_back(p);
}
pch = strtok(NULL, "/");
}
if (v.empty()) return "/";
string res;
for (int i = 0; i < v.size(); ++i) {
res += '/' + v[i];
}
return res;
}
};

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