您可以根据输入创建人员列表。它可以是一个对象，其中每个键是一个人的名字，对应的值是一个名字数组，代表那个人的朋友。当然你要确保当你将 B 添加为 A 的好友时，你也必须将 A 添加为 B 的好友。
对于示例输入，上述结构如下所示：
{
"Alden": ["Toshiko","Sharon","Che"],
"Toshiko": ["Alden"],
"Che": ["Kortney","Dorian","Alden"],
"Kortney": ["Che"],
"Dorian": ["Che","Quinn","Tyra"],
"Ronda": ["Lindell"],
"Lindell": ["Ronda"],
"Sharon": ["Alden"],
"Quinn": ["Dorian","Ally"],
"Owen": ["Sydnee"],
"Sydnee": ["Owen"],
"Tyra": ["Dorian"],
"Ally": ["Quinn"]
}
然后跟踪一个名字列表，从 Quinn 开始，还有一个距离，从 0 开始。
然后对于该列表中的每个名称，将当前距离指定为其 QDist 值。然后找到他们的朋友并将他们放在一起。删除已经收到 QDist 值的名称。
然后增加距离，并为新的名称列表重复上述操作。
不断重复，直到名称列表为空。
请注意，如果您按正确的顺序做事，您可以用 QDist 值替换好友列表。所以上面的结构在前两次迭代后会变成：
{
"Alden": ["Toshiko","Sharon","Che"],
"Toshiko": ["Alden"],
"Che": ["Kortney","Dorian","Alden"],
"Kortney": ["Che"],
"Dorian": 1,
"Ronda": ["Lindell"],
"Lindell": ["Ronda"],
"Sharon": ["Alden"],
"Quinn": 0,
"Owen": ["Sydnee"],
"Sydnee": ["Owen"],
"Tyra": ["Dorian"],
"Ally": 1
}
算法完成后，您有：
{
"Alden": 3,
"Toshiko": 4,
"Che": 2,
"Kortney": 3,
"Dorian": 1,
"Ronda": ["Lindell"],
"Lindell": ["Ronda"],
"Sharon": 4,
"Quinn": 0,
"Owen": ["Sydnee"],
"Sydnee": ["Owen"],
"Tyra": 2,
"Ally": 1
}
现在剩下的friends数组需要替换为“uncool”，因为显然对应的人与Quinn没有关系。列表也需要排序。
剧透警告！
这是一个有效的 sn-p：
// Get input as text
var input = `10
Alden Toshiko
Che Kortney
Che Dorian
Ronda Lindell
Sharon Alden
Dorian Quinn
Owen Sydnee
Alden Che
Dorian Tyra
Quinn Ally`;
// Build persons list with friends list
var persons =
// Take the input string
input
// Split it by any white-space to get array of words
.split(/\s+/)
// Skip the word at position 0: we don't need the line count
.slice(1)
// Loop over that array and build an object from it
.reduce(
// Arguments: obj = result from previous iteration
// name = current name in names array
// i = index in that array
// names = the whole array being looped over
(obj, name, i, names) => (
// Get the list of friends we already collected for this name.
// Create it as an empty array if not yet present.
obj[name] = (obj[name] || [])
// Add to that list the previous/next name, depending
// whether we are at an odd or even position in the names array
.concat([names[i%2 ? i-1 : i+1]])
// Use the updated object as return value for this iteration
, obj)
// Start the above loop with an empty object
, {});
// Now we have a nice object structure:
// { [name1]: [friendName1,friendName2,...], [name2]: ... }
// Start with Quinn as the only person we currently look at.
var friends = ['Quinn'];
// Increment the distance for each "generation" of friends
// until there are none left.
for (var i = 0; friends.length; i++) {
// Replace the friends list with a new list,
// while giving the friends in the current list a distance
friends =
// Start with the current list of friends
friends
// Loop over these friends.
// Only keep those that still have a friends array (object) assigned to them,
// since the others were already assigned a distance number.
.filter(friend => typeof persons[friend] === "object")
// Loop over those friends again, building a new list of friends
.reduce((friends, friend, k) => [
// Add this friends' friends to the new list
friends.concat(persons[friend]),
// ... and then replace this friends' friend list
// by the current distance we are at.
persons[friend] = i
// Return the first of the above two results (the new list)
// for the next iteration.
][0]
// Start with an empty array for the new friends list
, []);
}
// Now we have for each person that connects to Quinn a number:
// { [name1]: number, ... }
// Convert this to a format suitable to output
var result =
// Get list of names from the object (they are the keys)
Object.keys(persons)
// Sort that list of names
.sort()
// Loop over these names to format them
.map(name =>
// Format as "name: distance" or "name: uncool" depending on whether there
// still is an array of friends (object) in this entry
name + ': ' + (typeof persons[name] == 'object' ? 'uncool' : persons[name]));
// Output the result in the console
console.log(result);
还有一个更冗长但更容易理解的版本：
// Get input as text
var input = `10
Alden Toshiko
Che Kortney
Che Dorian
Ronda Lindell
Sharon Alden
Dorian Quinn
Owen Sydnee
Alden Che
Dorian Tyra
Quinn Ally`;
// Build persons list with friends list
// Take the input string
var persons = input;
// Split it by any white-space to get array of words
persons = persons.split(/\s+/)
// Skip the word at position 0: we don't need the line count
persons = persons.slice(1)
// Loop over that array and build an object from it
var obj = {}; // Start loop with an empty object
for (var i = 0; i < persons.length; i++) {
var name = persons[i]; // name = current name in names array
// Get the list of friends we already collected for this name.
// Create it as an empty array if not yet present.
if (obj[name] === undefined) obj[name] = [];
// Add to that list the previous/next name, depending
// whether we are at an odd or even position in the names array
obj[name].push(persons[i%2 === 1 ? i-1 : i+1]);
}
// Assign result to persons
persons = obj;
// Now we have a nice object structure:
// { [name1]: [friendName1,friendName2,...], [name2]: ... }
// Start with Quinn as the only person we currently look at.
var friends = ['Quinn'];
// Increment the distance for each "generation" of friends
// until there are none left.
for (var i = 0; friends.length !== 0; i++) {
// Loop over those friends, building a new list of friends
// Start with an empty array for the new friends list
var newFriends = [];
for (var k = 0; k < friends.length; k++) {
var friend = friends[k];
// Only consider those that still have a friends array (object) assigned to them,
// since the others were already assigned a distance number.
if (typeof persons[friend] === "object") {
// Add this friends' friends to the new list
newFriends = newFriends.concat(persons[friend]);
// ... and then replace this friends' friend list
// by the current distance we are at.
persons[friend] = i;
}
};
// Make the new list the current list:
friends = newFriends;
}
// Now we have for each person that connects to Quinn a number:
// { [name1]: number, ... }
// Convert this to a format suitable to output
// Get list of names from the object (they are the keys)
var result = Object.keys(persons);
// Sort that list of names
result.sort();
// Loop over these names to format them
for (var i = 0; i < result.length; i++) {
var name = result[i];
// Format as "name: distance" or "name: uncool" depending on whether there
// still is an array of friends (object) in this entry
if (typeof persons[name] == 'object') {
result[i] = name + ': uncool';
} else {
result[i] = name + ': ' + persons[name];
}
}
// Output the result in the console
console.log(result);