Simplify Path


Given an absolute path for a file (Unix-style), simplify it.
For example,
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".
Solution: Use stack to deal with "..". I use the approach in section 7.3 of this page to split the string into tokens for given delimiter.

Another approach is to first copy the string into a c_str, then use strtok.
Or we can directly traverse through the string and check the delimiter. There is no essential difference.

Comments

Post a Comment

Popular posts from this blog

House Robber

You are a professional robber planning to rob houses along a street. Each house has a certain amount of money stashed, the only constraint stopping you from robbing each of them is that adjacent houses have security system connected and it will automatically contact the police if two adjacent houses were broken into on the same night.  Given a list of non-negative integers representing the amount of money of each house, determine the maximum amount of money you can rob tonight without alerting the police. class Solution { public: int rob(vector &num) { if (num.size() < 1) return 0; vector result; result.push_back(num[0]); for (int i = 1; i < num.size(); ++i) { int curr_max = 0; for (int j = 0; j < i - 1; ++j) { curr_max = max(curr_max, num[i] + result[j]); } curr_max = max(curr_max, result[i - 1]); curr_max = max(curr_max, num[i]); result.p...
Read more

Maximum Gap

Given an unsorted array, find the maximum difference between the successive elements in its sorted form.  Try to solve it in linear time/space.  Return 0 if the array contains less than 2 elements. You may assume all elements in the array are non-negative integers and fit in the 32-bit signed integer range. In order to solve this in linear time, we cannot use general sorting algorithm. Here we use bucket sort. First, we scan the array to find the min and the max values.   We can then create n - 1 buckets with equal size. Suppose we have n elements in total, then we have n - 2 elements except the min and the max. Scan the array again to put the n - 2 elements into these buckets. We should have at least one empty bucket. For each bucket, we record the min and the max for all the elements within that bucket.  Finally, we scan all the buckets, and maintain the maximum gap. Note that the maximum gap between the successive elements in the sorted...
Read more

[ITint5] Maximum Subarray for a Circular Array

http://www.itint5.com/oj/#8 Classic DP problem, which has also appeared in Leetcode as Maximum subarray . The only difference is, in this problem, empty subarray (where sum is 0) is allowed. So instead of return ret, we could simply return max(0, ret). http://www.itint5.com/oj/#9 Calculate the maximum subarray for a circular array. For example, [1, 3, -2, 6, -1], we should return 9 since 6 + (-1) + 1 + 3 = 9. The following solution might not be optimized. Suppose that the maximum subarray is from A[i] to A[j]. The idea is, consider the following different scenarios: i <= j: it is the same as classic problem. i > j: the sum is A[i : end] + A[start : j]. So we can use two additional arrays: left[j] = maximum subarray, starts from A[start] and ends no later than A[j] right[i] = maximum subarray, starts no earlier than A[i] and ends at A[end] The following code passes the online judge. However, we need O(4n) in time and O(2n) in space. Not sure how to improve this. ...
Read more