BASIC Javascript 数组函数，问题是已知的，但我无法理解解决方案

Question

In the below function I am attempting to get an output which resembles this:

[[1,1,1,1],[2,2,2], 4,5,10,[20,20], 391, 392,591].

I can see that the problem I have embedded is that I am always adding the temp array with a push to the functions return, as a result, all of the individual numbers apart from the last number in the for each function are being pushed into the target array with the array object also.

I feel as though I need a further conditonal check but for the life of me I am unable to come up with solution which works.

Any suggestions would be much appreciated.

const sortme = (unsortedArr)=> {

let tempArr = []; 

let outputArr = []; 

const reorderedArr = unsortedArr.sort((a,b) => a-b); 

reorderedArr.forEach((number, i) => {

    if ((i === 0) || (reorderedArr[i] === reorderedArr[i-1])) {
    tempArr.push(number);  
    }

    else {     
    outputArr.push(tempArr);
    tempArr = [];
    tempArr.push(number); 
}  
})

outputArr.push(tempArr[0]);
    return outputArr; 
}

const unsortedArr = [1,2,4,591,392,391,2,5,10,2,1,1,1,20,20]; 

sortme(unsortedArr); 

Answer 1

i would make a deduped copy and .map() it to transform the values into arrays containing values from the original ( sorted ) array that you get using a .forEach :

 const unsortedArr = [1, 2, 4, 591, 392, 391, 2, 5, 10, 2, 1, 1, 1, 20, 20]; const sortMe = (arr) => { arr = arr.sort((a, b) => a - b); // a short way to dedupe an array // results in : 1, 2, 4, 5, 10, 20, 391, 392, 591 let dedupe = [...new Set(arr)]; let tmpArr; return dedupe.map(e => { tmpArr = []; // empty tmpArr on each iteration // for each element of the deduped array, look for matching elements in the original one and push them in the tmpArr arr.forEach(a => { if (a === e) tmpArr.push(e); }) if(tmpArr.length === 1) return tmpArr[0]; // in case you have [4] , just return the 4 else return tmpArr; // in case you have [1,1,1,1] // shorthand for the if/else above // return tmpArr.length === 1 ? tmpArr[0] : tmpArr; }); } const result = sortMe(unsortedArr); console.log(result);

Answer 2

This should work (using reduce):

 const unsortedArr = [1,2,4,591,392,391,2,5,10,2,1,1,1,20,20]; let lastValue = null; var newArr = unsortedArr.sort((a,b) => ab).reduce((acc, value) => { if (acc.length == 0 || ((acc.length > 0 || !acc[acc.length-1].length) && lastValue !== value)) { acc.push(value); } else if (acc.length > 0 && lastValue === value) { acc[acc.length-1] = (acc[acc.length-1].length ? acc[acc.length-1].concat([value]): [value, value]); } lastValue = value; return acc; }, []); console.log(newArr);

Answer 3

And another approach, just for fun:

 const unsortedArr = [1,2,4,591,392,391,2,5,10,2,1,1,1,20,20]; var arr = unsortedArr.sort((a,b) => ab).reduce((acc, value) => { if (acc.length > 0 && acc[acc.length-1].includes(value)) { acc[acc.length-1].push(value); } else { acc.push([value]) } return acc; }, []).map((v) => v.length > 1 ? v: v[0]); console.log(arr);

Answer 4

I hope the below one is quite simple;

 function findSame(pos, sortedArr){ for(let i =pos; i<sortedArr.length; i++){ if(sortedArr[i] !== sortedArr[pos]){ return i } } } function clubSameNumbers(unsortedArr){ let sortedArr = unsortedArr.sort((a,b)=>ab) //[ 1, 1, 1, 1, 2, 2, 2, 4, 5, 10, 20, 20, 391, 392, 591 ] let result = [] for(let i = 0; i < sortedArr.length; i = end){ let start = i var end = findSame(i, sortedArr) let arr = sortedArr.slice(i, end) arr.length > 1 ? result.push(arr) : result.push(...arr) } return result } console.log(clubSameNumbers([1,2,4,591,392,391,2,5,10,2,1,1,1,20,20])) //[ [ 1, 1, 1, 1 ], [ 2, 2, 2 ], 4, 5, 10, [ 20, 20 ], 391, 392, 591 ]