如何在多维数组中找到彼此相邻的相同值？

Question

I tried to write a function that would find in a multidimensional array (with values from 3 to 7) repeating values for at least 3 times next to each other (vertical and horizontal). And if it finds that, change it for a different value. Let's say 1.

I tried to do this by loops but it doesn't seem to be a good way to solve that or I messed it up. Because for some array it works, for some it does not.

Here's my code:

 function searching(array) { for (i = 0; i < array.length; i++) { let horizontal = array[i][0]; let howMany = 1; for (j = 1; j < array[i].length; j++) { if (horizontal === array[i][j]) { howMany += 1; horizontal = array[i][j]; if (howMany >= 3) { for (d = j; d > j - howMany; d--) { array[i][d] = 0; } } } else { horizontal = array[i][j]; howMany = 1; } } } for (v = 0; v < array.length; v++) { let vertical = array[0][v]; let howMany = 1; for (x = 1; x < array.length; x++) { if (vertical === array[x][v]) { howMany++; vertical = array[x][v]; if (howMany >= 3) { for (d = x; d > x - howMany; d--) { array[d][v] = 0; } } } else { vertical = array[x][v]; howMany = 1; } } } } 

The idea is to for example give array:

 let array = [ [3, 4, 5, 6, 7], [3, 4, 5, 6, 7], [3, 4, 5, 5, 5], [3, 5, 6, 7, 4] ] 

And the result should be:

 let result = [ [1, 1, 1, 6, 7], [1, 1, 1, 6, 7], [1, 1, 1, 1, 1], [1, 5, 6, 7, 4] ] 

Thanks in advance for any ideas how to solve it :) Greetings!

Answer 1

I didn't understand the question at first... So this is my code:

 let array = [ [3, 4, 5, 6, 7], [3, 4, 5, 6, 7], [3, 4, 5, 5, 5], [3, 5, 6, 7, 4] ]; function replace(arr, target = 1) { let needToChange = []; // save the index to change const numbers = [3, 4, 5, 6, 7]; const m = arr.length; // m rows const n = arr[0].length; // n columns let mi = 0; let ni = 0; // search in row for (mi = 0; mi < m; mi++) { for (let x = 0; x < numbers.length; x++) { const num = numbers[x]; // number to search let counter = 0; // counter for this number in row mi let tempArr = []; for (ni = 0; ni < n; ni++) { const currentNum = arr[mi][ni]; if (currentNum === num) { counter++; tempArr.push([mi, ni]); } } if (counter >= 3) { needToChange = needToChange.concat(tempArr); } } } // search in column for (ni = 0; ni < n; ni++) { for (let x = 0; x < numbers.length; x++) { const num = numbers[x]; // number to search let counter = 0; // counter for this number in row mi let tempArr = []; for (mi = 0; mi < m; mi++) { const currentNum = arr[mi][ni]; if (currentNum === num) { counter++; tempArr.push([mi, ni]); } } if (counter >= 3) { needToChange = needToChange.concat(tempArr); } } } // replace needToChange.forEach(([i, j]) => { array[i][j] = target; }); } replace(array); array.forEach(row => { console.log(row.join(', ')); }) 

Answer 2

The problems with your current code are

(1) You're only checking individual rows and columns, when you need to be checking them both (eg, with [[2, 2], [2, 5]] , when at the starting position [0][0] , you need to look at both [0][1] (and its neighbors, if matching) as well as [1][0] (and its neighbors, if matching).

(2) You're not actually checking for adjacency at the moment, you're just counting up the total number of matching elements in a particular row or column.

Iterate over all indicies of the array. If an index has already been checked, return early. Recursively search for neighbors to that index, and if at least 3 matching in total are found, set them all to 1. Put all matching neighbors in the checked set to avoid checking them again (even if there were less than 2 adjacent matches found total).

 setAllAdjacentToOne([ [3, 4, 5, 6, 7], [3, 4, 5, 6, 7], [3, 4, 5, 5, 5], [3, 5, 6, 7, 4] ]); // all 9s stay, the rest get set to 1: setAllAdjacentToOne([ [2, 2, 9, 7, 7], [2, 9, 9, 9, 7], [3, 4, 4, 5, 5], [9, 4, 5, 5, 9] ]); function setAllAdjacentToOne(input) { const output = input.map(subarr => subarr.slice()); const checked = new Set(); const getKey = (x, y) => `${x}_${y}`; const width = input[0].length; const height = input.length; const getAllAdjacent = (x, y, numToFind, matches = []) => { if (x >= width || x < 0 || y >= height || y < 0) { return matches; } const key = getKey(x, y); if (!checked.has(key) && input[y][x] === numToFind) { checked.add(key); matches.push({ x, y }); getAllAdjacent(x + 1, y, numToFind, matches); getAllAdjacent(x - 1, y, numToFind, matches); getAllAdjacent(x, y + 1, numToFind, matches); getAllAdjacent(x, y - 1, numToFind, matches); } return matches; }; output.forEach((innerRowArr, y) => { innerRowArr.forEach((num, x) => { const allAdjacent = getAllAdjacent(x, y, num); if (allAdjacent.length <= 2) { return; } allAdjacent.forEach(({ x, y }) => { output[y][x] = 1; }); }); }); console.log(JSON.stringify(output)); }