CONY的世界

71. 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.

這題我是用 split 和 join 來做，不過看起來有點慢，但是大致想法是對的
因為程式碼很短執行時間也很短，所以只要寫法有點繞就會慢，附上別人寫的比較精簡的寫法

Success:

62 ms, beats 20.17 % of python submissions 

class Solution(object):
    def simplifyPath(self, path):
        """
        :type path: str
        :rtype: str
        """
        result = []
        stack = path.split("/")
        for index, item in enumerate(stack):
            if not item or item == ".": continue
            if item == "..":
                if result: result.pop()
                continue
            result.append((stack[index]))

        result = "/" + "/".join(result)
        return result

較快的寫法

42 ms, beats 72.88 % of python submissions 

class Solution(object):
    def simplifyPath(self, path):
        places = [p for p in path.split("/") if p!="." and p!=""]
        stack = []
        for p in places:
            if p == "..":
                if len(stack) > 0:
                    stack.pop()
            else:
                stack.append(p)
        return "/" + "/".join(stack)

402. Remove K Digits

Input: num = "1432219", k = 3
Output: "1219"
Explanation: Remove the three digits 4, 3, and 2 to form the new number 1219 which is the smallest.

Input: num = "10200", k = 1
Output: "200"
Explanation: Remove the leading 1 and the number is 200. Note that the output must not contain leading zeroes.

Input: num = "10", k = 2
Output: "0"
Explanation: Remove all the digits from the number and it is left with nothing which is 0.

這題一開始想得太複雜，想說要找出min 並包含之後的數值
最後看了大家的做法，是用剔除會讓值變大的方式來做 （與之前值比較，如果之前值較大就pop 踢出）

注意一下因為可能會有 "9" 1 -> "0" 這種case，所以要在最後加上 out[:-k or None]

out[:-k] 來解決k 可能沒歸0的可能，一般來說是走 out[:None] ＝ out[:] 也就是全印的情況

然後跟最後可能沒東西的狀況，所以還要  or "0"

Success:

65 ms, beats 57.54 % of python submissions 

class Solution(object):

    def removeKdigits(self, num, k):
        """
        :type num: str
        :type k: int
        :rtype: str
        """
        out = []
        for n in num:
            while k and out and n < out[-1]:
                    out.pop()
                    k -= 1
            out.append(n)
        return "".join(out[:-k or None]).lstrip("0") or "0"