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This article mainly explains "how to solve the problem of train ticket sorting by js". Interested friends may wish to have a look. The method introduced in this paper is simple, fast and practical. Let's let the editor take you to learn "how to solve the problem of train ticket ordering by js".

Topic: a person sets out from Beijing and travels by train through N cities to Shanghai. How to rank them when they get these train tickets.

Analyze the description of this topic, the essence is array sorting, but this array is special, each item of the array should be an object, the object has two properties, that is, the starting station and the terminal of each ticket.

So the first step should be to build this data structure, as follows:

Var arr = [

{start: 'Suning', end: 'Changsha'}

{start: 'Cangzhou', end: 'Renqiu'}

{start: 'Renqiu', end: 'Suning'}

{start: 'Changsha', end: 'Wuhan'}

{start: 'Wuhan', end: 'Shanghai'}

{start: 'Beijing', end: 'Cangzhou'}

]

Now that the data structure is simulated, the next step is to arrange it in order. The idea of solving the problem is to first find out the ticket at the starting station of the trip, and then cycle through the array. According to the fact that the terminal of the current item is the same as the starting station of the next ticket, find out the next train ticket and arrange it in order.

How to use the code to find the first ticket? You can take a closer look at this data and find the following characteristics: the starting station of the first ticket will never appear in the terminal. Use this feature to find out the ticket at the starting station. The code is as follows (see notes for details):

/ / calculate the starting station of each train

Var startarr= arr.map ((ePermil) = > {

Return e.start

});

/ / calculate the terminal of each train

Var endarr = arr.map ((ePermil) = > {

Return e.end

});

/ / define a new array

Var arr2 = []

/ / the starting station that uses the purpose of the trip will never appear in the terminal. Filter it out and fill it into the array.

For (var iTuno Tinci {

Return e.start = = startarr [I]

})

Arr2.push (a [0])

}

}

After that, loop through the disordered ticket array, and take out the last item of the arr2 array each time. According to the end phase of this item, query the tickets with the same start attribute in the out-of-order array. After finding them, push them into the arr2 array and loop through them to get the sorted array. The code is as follows:

For (var iTuno Tinci {

Return e.start = = ticket.end

});

If (tt.length==1) {

Arr2.push (tt [0])

}

}

Looking back at the code today, of course, it is a little immature, so I have optimized the above code with some features of se6, as follows:

Var arr = [

{start: 'Suning', end: 'Changsha'}

{start: 'Cangzhou', end: 'Renqiu'}

{start: 'Renqiu', end: 'Suning'}

{start: 'Changsha', end: 'Wuhan'}

{start: 'Wuhan', end: 'Shanghai'}

{start: 'Beijing', end: 'Cangzhou'}

]

Let a = arr.filter ((ePermie iMagol arr) = > arr.map (v = > v.end) .indexOf (e.start) = =-1)

Let b = arr.reduce ((pre,cur,index,arr) = > {

Return pre.concat (arr.filter ((e) I) = > {

Return e.start = = pre.length-1] .end

}))

}, a)

Console.log (b)

The method of removing the weight of the array:

Function unique (arr) {

Return Array.from (new Set (arr))

}

Function unique (arr) {

Return [. New Set (arr)]

}

Function unique (arr) {

Var obj = {}

Return arr.filter (e = > {return! obj [e] & & (obj [e] = 1)})

} at this point, I believe you have a deeper understanding of "how to solve the problem of train ticket sorting by js". You might as well do it in practice. Here is the website, more related content can enter the relevant channels to inquire, follow us, continue to learn!

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