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✨ topic

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The question has beenForce button brush column included!

1436. Tour terminus

Paths [I] = [cityAi, cityBi] indicates that the paths will go directly from cityAi to cityBi. Find the final destination of the tour, a city that does not have any route to any other city.

The problem data guarantees that the route diagram will form a circuit with no loops, so that there is exactly one travel destination.

Example 1:

Enter: Paths = [[“London”,”New York”],[“New York”,”Lima”],[“Lima”,”Sao Paulo”] It starts at London and ends at Sao Paulo. The itinerary for this trip is “London” -> “New York” -> “Lima” -> “Sao Paulo”.

Example 2:

Input: paths = [[” B “, “C”], [” D “, “B”], [” C “, “A”]] output: “A” explanation: all possible routes are: “D” -> “B” -> “C” -> “A”. “B” -> “C” -> “A”. “C” -> “A”. “A”. Obviously, the end of the trip is “A”.

Example 3:

Paths = [[“A”,”Z”]]

Tip:

1 <= paths.length <= 100 paths[i].length == 2 1 <= cityAi.length, cityBi.length <= 10 cityAi ! = cityBi All characters consist of uppercase and lowercase letters and Spaces.


🔥 : hash table

They already say:A city without any lines to other cities.That is to sayThere is only one key station:Starting from any of these stations can lead to the final site

So we use hash table simulation:K:V -- > Current site: next site, so we can keep all the mappings, so we can start at any site

🔥 If the current site does not have another site, the site is an important site.


✨ code implementation

class Solution {
    public String destCity(List<List<String>> paths) {

        Map<String, String> map = new HashMap<>();

        int len = paths.size();
        
        for (int i = 0; i < len; i++) {
            map.put(paths.get(i).get(0),paths.get(i).get(1));
        }
        
        String value = paths.get(0).get(1);
        
        while (map.containsKey(value)) {
            value = map.get(value);
        }

        returnvalue; }}Copy the code

💖 finally

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