Inserting elements in a matrix spirally

问题:

Given a number x, insert elements 1 to x^2 in a matrix spirally. e.g. For x = 3, matrix looks like [[1,2,3],[8,9,4],[7,6,5]]. For this I've written following snippet. However, I'm getting o/p as [[7,9,5],[7,9,5],[7,9,5]]

while(t<=b && l<=r){
               System.out.print(t+" "+b+" "+l+" "+r+"\n");
        if(dir==0){
            for(int i = l;i<=r;i++){
                arr.get(t).set(i,x);
                x++;
            }

            t++;
        }else if(dir==1){
            for(int i = t;i<=b;i++){
                arr.get(i).set(r,x);
                x++;
            }
            r--;
        }else if(dir==2){
            for(int i = r;i>=l;i--){
                arr.get(b).set(i,x);
                x++;
            }
            b--;
        }else if(dir==3){
            for(int i = b;i>=t;i--){
                arr.get(l).set(i,x);
                x++;
            }
            l++;
        }
        dir = (dir+1)%4;

    }

回答1:

You can use the next code (which I developed for some implementation that handles huge martrix sizes). It will use width (columns) and height (rows) of any matrix size and produce the output you need

    List<rec> BuildSpiralIndexList(long w, long h)
    {
        List<rec> result = new List<rec>();
        long count = 0,dir = 1,phase = 0,pos = 0;
        long length = 0,totallength = 0;
        bool isVertical = false;

        if ((w * h)<1) return null;
        do
        {
            isVertical = (count % 2) != 0;
            length = (isVertical ? h : w) - count / 2 - count % 2;
            phase = (count / 4);
            pos = (count % 4);
            dir = pos > 1 ? -1 : 1;
            for (int t = 0; t < length; t++)
                // you can replace the next code with printing or any other action you need
                result.Add(new rec()
                {
                    X = ((pos == 2 || pos == 1) ? (w - 1 - phase - (pos == 2 ? 1 : 0)) : phase) + dir * (isVertical ? 0 : t),
                    Y = ((pos <= 1 ? phase + pos : (h - 1) - phase - pos / 3)) + dir * (isVertical ? t : 0),
                    Index = totallength + t
                });
            totallength += length;
            count++;
        } while (totallength < (w*h));
        return result;
    }

回答2:

This solution walks from the top left to the top right, the top right to the bottom right, the bottom right to the bottom left and the bottom left up to the top left. It is a tricky problem, hopefully my comments below assist in explaining.

Below is a codepen link to see it added to a table. https://codepen.io/mitchell-boland/pen/rqdWPO

const n = 3; // Set this to a number

matrixSpiral(n);

function matrixSpiral(number){

    // Will populate the outer array with n-times inner arrays
    var outerArray = [];

    for(var i = 0; i < number; i++){
      outerArray.push([]);
    }


    var leftColumn = 0;
    var rightColumn = number - 1;
    var topRow = 0;
    var bottomRow = number-1;
    var counter = 1; // Used to track the number we are up to.

    while(leftColumn <= rightColumn && topRow  <=bottomRow){

        // populate the top row
        for(var i = leftColumn; i <= rightColumn; i++){
          outerArray[leftColumn][i] = counter;
          counter++;
        }
        // Top row is now populated
        topRow ++;

        // Populate the right column
        for(var i = topRow ; i <= bottomRow; i++){
          outerArray[i][rightColumn] = counter;
          counter++;
        }
        // Right column now populated.
        rightColumn--;

        // Populate the bottom row
        // We are going from the bottom right, to the bottom left
        for(var i = rightColumn; i >= leftColumn; i--){
          outerArray[bottomRow][i] = counter;
          counter++;
        }
        // Bottom Row now populated
        bottomRow--;

        // Populate the left column
        // We are going from bottom left, to top left
        for(var i = bottomRow; i >= topRow ; i--){
          outerArray[i][leftColumn] = counter;
          counter++;
        }
        // Left column now populated.
        leftColumn++;

        // While loop will now repeat the above process, but a step in.
    }

    // Console log the results.
    for(var i = 0; i < number; i++){
        console.log(outerArray[i]);
    }
}