JavaScript implements listing the longest consecutive number in an array_javascript skills

Original title:

Given an unordered sequence of integers, find the longest sequence of consecutive numbers.

For example:

Given [100, 4, 200, 1, 3, 2],

The longest continuous sequence of numbers is [1, 2, 3, 4].

The solution given by Xiao Cai:

Copy code The code is as follows:

function maxSequence(array,step){
var _array = array.slice(), //clone array
_step = 1,
​ _arrayTemp = [],
i = 0;
var parseLogic = {
//result container
ParseResults: [],
//set value to array,what's the last array of parseResults
set: function(n){
This.parseResults[this.parseResults.length-1].push(n);
},
//get the last array from parseResults
Get: function(){
Return this.parseResults[this.parseResults.length-1];
},
//put a new array in parseResults
AddItem: function(){
This.parseResults.push([]);
},
//sort parseResults
SortByAsc: function(){
This.parseResults.sort(function(a,b){
           return a.length - b.length;
});
}
};
//check params
_step = step || _step;
//sort array by asc
_array.sort(function(a,b){
Return a - b;
});
//remove repeat of data
for(i = 0;i<_array.length;i ){
If(_array[i] != _array[i 1]){
​ _arrayTemp.push(_array[i]);
}
}
_array = _arrayTemp.slice();
_arrayTemp = [];
//parse array
parseLogic.addItem();
for(i = 0;i<_array.length;i ){
If(_array[i] _step == _array[i 1]){
       parseLogic.set(_array[i]);
Continue;
}
If(_array[i]-_step == _array[i-1]){
       parseLogic.set(_array[i]);
       parseLogic.addItem();
}
}
//sort result
parseLogic.sortByAsc();
//get the max sequence
Return parseLogic.get();
}

Calling instructions:

Method name:

maxSequence(array,step)

Parameter description:

Array: The array to be searched. necessary.

step: sequence step (increment). Optional, default is 1.

Return value:

This method will not change the passed array, but will return a new array containing the largest sequence.

Call example:

maxSequence([5,7,2,4,0,3,9],1); //return [2,3,4,5]

maxSequence([5,7,2,4,0,3,9],2); //return [5,7,9]