LeetCode Brush Questions 1436. Travel Terminal

LeetCode Brush Questions 1436. Travel Terminal

I don't know where I am going, but I am already on my way!
Time is hurried, although I have never met, but I met Yusi, it is really a great fate, thank you for your visit!

• Topic :
  give you a tour circuit diagram, the circuit diagram of an array of travel routes paths, where paths[i] = [cityAi, cityBi]representation from the line will cityAigo directly cityBi. Please find out the destination of this trip, that is, a city that does not have any routes to other cities.
  The title data guarantees that the route map will form a route without loops, so there will only be one travel terminal.
• Example :

示例 1 :
输入：paths = [["London","New York"],["New York","Lima"],["Lima","Sao Paulo"]]
输出："Sao Paulo" 
解释：从 "London" 出发，最后抵达终点站 "Sao Paulo" 。本次旅行的路线是 "London" -> "New York" -> "Lima" -> "Sao Paulo" 。

示例 2 :
输入：paths = [["B","C"],["D","B"],["C","A"]]
输出："A"
解释：所有可能的线路是：
"D" -> "B" -> "C" -> "A". 
"B" -> "C" -> "A". 
"C" -> "A". 
"A". 
显然，旅行终点站是 "A" 。

示例 3 :
输入：paths = [["A","Z"]]
输出："Z"

• Tips :
  • 1 <= paths.length <= 100
  • paths[i].length == 2
  • 1 <= cityAi.length, cityBi.length <= 10
  • cityAi != cityBi
  • All strings are composed of uppercase and lowercase English letters and space characters.
• Code 1:

class Solution:
    def destCity(self, paths: List[List[str]]) -> str:
        cityAi = []
        cityBi = []
        for i in range(len(paths)):
            cityAi.append(paths[i][0])
            cityBi.append(paths[i][1])
        for i in range(len(paths)):
            if cityBi[i] not in cityAi:
                return cityBi[i]
# 执行用时 :36 ms, 在所有 Python3 提交中击败了94.14%的用户
# 内存消耗 :13.7 MB, 在所有 Python3 提交中击败了100.00%的用户

• Algorithm description:
  Find the starting point and the end point respectively, and judge whether the end point appears in the starting point, if not, return to the end point.
• Code 2:

class Solution(object):
    def destCity(self, paths: List[List[str]]) -> str:
        citys = set()
        cityBi = set()
        for path in paths:
            citys.add(path[0])
            citys.add(path[1])
            cityBi.add(path[0])
        return (citys - cityBi).pop()
# 执行用时 :40 ms, 在所有 Python3 提交中击败了82.03%的用户
# 内存消耗 :13.7 MB, 在所有 Python3 提交中击败了100.00%的用户

• Algorithm description:
  Using the set nature, subtract the set of starting point cities from the set of all cities, and the remaining cities are the ending cities.