javascript - find sequence of numbers in array and alert highest number - Stack Overflow

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I have an array that looks like this:

[1,2,3,4,5,6,8,10,12,13,14,15,20,21,22,23,24,25,26,27,28,29,30]

the highest count on one of the found sequences would be: 10

My goal is to loop through the array and identify the sequences of numbers, then find the length of the highest sequence that exists.

So, based on the array above, the length of the longest sequence would be "10"

Does anyone know of quick and easy script to find this?

I have an array that looks like this:

[1,2,3,4,5,6,8,10,12,13,14,15,20,21,22,23,24,25,26,27,28,29,30]

the highest count on one of the found sequences would be: 10

My goal is to loop through the array and identify the sequences of numbers, then find the length of the highest sequence that exists.

So, based on the array above, the length of the longest sequence would be "10"

Does anyone know of quick and easy script to find this?

• javascript

Share Improve this question Follow edited Jun 13, 2014 at 23:46 Web_Designer 74.7k9393 gold badges210210 silver badges267267 bronze badges asked Jul 20, 2011 at 21:39 jakejake 31755 silver badges1919 bronze badges 4

• 1 Isn't the length of the longest sequence 11 (20-30)? – carlosfigueira Commented Jul 20, 2011 at 21:45
• 1 This sounds like homework. Have you attempted anything yet? – Luke Commented Jul 20, 2011 at 21:50
• Is 2,4,6 classed as 1 sequence? – Ash Burlaczenko Commented Jul 20, 2011 at 21:56
• BTW, are you going to accept anything as answer? ;) – Mo Valipour Commented Jul 20, 2011 at 22:09

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3 Answers 3

Sorted by: Reset to default 4

OK, I think I found a very short way of doing this (only 1 line for the for loop):

var arr = [1,2,3,4,5,6,8,10,12,13,14,15,20,21,22,23,24,25,26,27,28,29,30];
var res = new Array();
res[0] = 0;

for(var i=1;i<arr.length;i++) res[i] = (arr[i] == arr[i-1] + 1) ? (res[i-1] + 1) : 0;

var maxLength = Math.max.apply({},res);

this gives you (10) as the result. if you need (11) (which makes more sense) change the 0 to 1 in the for loop.

jsFiddle link: http://jsfiddle/gEzzA/8/

You don't need jQuery for this.

function longestSeq(arr) {
  var len = 0, longestLen = -1, prev = null;

  for (var i = 0; i < arr.length; ++i) {
    if (prev == null || arr[i] - 1 === prev)
      ++len;
    else {
      if (len > longestLen) longestLen = len;
      len = 1;
    }
  }
  return longestLen > len ? longestLen : len;
}

What that does is keep track of how long it's been since a "break" has been seen. Each time a break is seen, it checks whether the longest so far is shorter than the last good run.

Here's the solution in pseudo code...

First, setup another array with the same number of elements and initialised to zero, to use as counters...

Array01:=[1,2,3,4,5,6,8,10,12,13,14,15,20,21,22,23,24,25,26,27,28,29,30]
Array02:=[0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0]

Now the logic for filling in the counters...

FOR i:=0 TO LastElement DO 
 WHILE (Array01[i+1]-Array01[i]=1) AND (i<LastElement) DO Inc(Array02[i]);

Now to scan who's got the highest sequence score...

which:=0; Value:=Array02[0];
FOR i:=0 TO LastElement DO 
 IF Array02[i]>Value THEN BEGIN Value:=Array02[i]; Which:=i; END;

So, at the end of this the highest sequence is held by Array element "Which" and the count is "Value"!

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