71. Simplify Path
2017-10-16 本文已影响0人
Al73r
题目
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"
.
分析
字符串处理,就是记录每两个“/”之间的字符(跳过连续的“/”)。如果是“.”则不管。如果是正常字母则加入到答案中,并在前面加上“/”。如果是“..”则在答案中删去上一个“/”之后的字符。
实现
class Solution {
public:
string simplifyPath(string path) {
string ans;
int i=0;
while(i<path.size()){
while(i<path.size() && path[i]=='/') i++;
string tmp;
while(i<path.size() && path[i]!='/'){
tmp += path[i];
i++;
}
if(tmp=="." || tmp.empty()) continue;
if(tmp==".."){
if(!ans.empty()){
int j=ans.size()-1;
while(j>=0 && ans[j]!='/') j--;
ans.erase(ans.begin()+j, ans.end());
}
}
else{
ans += '/';
ans += tmp;
}
}
if(ans.empty()) ans += '/';
return ans;
}
};
思考
字符串处理特点就是繁杂,大家似乎都很不喜欢这种题,大家的对这种题的评价都不行。