Simplify Path

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"

.

模拟栈

/*
1.遇到./，当前位置-2,
2.遇到../,当前位置-3，然后回退到上一个斜杠处
测试用例：
.
..
/.
/..
/...
/./
/../
///
/./
/../
/.../
/..../../../../../../
/a../././../
/a.../././..
*/
class Solution {
public:
string simplifyPath(string path) {
int i=0,j=0;
if(path.size()>0){
path.push_back('/');
}
string res(path.size(),' ');
while(i<path.size()){
if(i>0 && path[i]==res[j-1] && path[i]=='/') {
i++;
continue;
}
res[j++] = path[i];
if(path[i]=='/'){
int pre_level_start = i-3>=0? i-3:0;
if(path.substr(pre_level_start,i-pre_level_start+1)=="/../"){
j-=4;
while(j-1>=0 && res[j-1]!='/') --j;
if(j==0){
j=1;
}
}else{
pre_level_start = i-2>=0? i-2:0;
if(path.substr(pre_level_start,i-pre_level_start+1)=="/./"){
j-=2;
}
}
}
i++;
}
if(j>1 && res[j-1]=='/') --j;
res.erase(res.begin()+j,res.end());
return res;
}
};

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