【LeetCode】52. Simplify Path

题目描述(Medium)

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

题目链接

https://leetcode.com/problems/simplify-path/description/

Example 1:

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

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".

算法分析

如果“..”则出栈，其他情况进栈

提交代码：

class Solution {
public:
	string simplifyPath(string path) {
        vector<string> dirs;
        
        for(auto i = path.begin(); i != path.end(); )
        {
            ++i;
            
            auto j = find(i, path.end(), '/');
            string dir = string(i, j);
            if(!dir.empty() && dir != ".")
            {
                if(dir == "..")
                {
                    if(!dirs.empty())
                        dirs.pop_back();
                }
                else
                    dirs.push_back(dir);
            }
            
            i = j;
        }
        
        stringstream out;
        if(dirs.empty())
            out << '/';
        else
        {
            for(string dir : dirs)
                out << '/' << dir;
        }
        
        return out.str();
	}
};

测试代码：

// ====================测试代码====================
void Test(const char* testName, string strs, string expected)
{
	if (testName != nullptr)
		printf("%s begins: \n", testName);

	Solution s;
	string result = s.simplifyPath(strs);

	if(result == expected)
		printf("passed\n");
	else
		printf("failed\n");
}

int main(int argc, char* argv[])
{
	
	Test("Test1", string("/home/"), string("/home"));
	Test("Test2", string("/a/./b/../../c/"), string("/c"));
	Test("Test3", string("/../"), string("/"));
	Test("Test4", string("/home//foo/"), string("/home/foo"));

	return 0;
}