I'm new to JavaScript, I'm trying to solve leetcode question 37. I need to a create a blank two dimensional array, I initially used the method in the comments; however, it doesn't work correctly, it will change all the value. Then, I used the for loop method to create array and currently it worked correctly. But I still cannot figured out why this will happen, could anyone explain the reason why this will happen, is this because of shallow copy?
var solveSudoku = function (board) {
// let rows = new Array(9).fill(new Array(10).fill(0)),
let rows = new Array(9);
for (let i = 0; i < 9; i++) {
rows[i] = new Array(10).fill(0);
}
let cols = new Array(9);
for (let i = 0; i < 9; i++) {
cols[i] = new Array(10).fill(0);
}
let boxes = new Array(9);
for (let i = 0; i < 9; i++) {
boxes[i] = new Array(10).fill(0);
}
// let cols = new Array(9).fill(new Array(10).fill(0)),
// boxes = new Array(9).fill(new Array(10).fill(0));
for (let i = 0; i < 9; i++) {
for (let j = 0; j < 9; j++) {
let c = board[i][j];
if (c !== '.') {
let n = parseInt(c),
bx = Math.floor(j / 3),
by = Math.floor(i / 3);
// 0代表为使用，1为使用过
rows[i][n] = 1;
console.log(i, n)
cols[j][n] = 1;
// box索引
boxes[by * 3 + bx][n] = 1;
}
}
}
fill(board, 0, 0)
function fill(board, x, y) {
// 完成填充条件
if (y === 9) return true;
// 下一个点的坐标
let nx = (x + 1) % 9,
// 判断进入是否下一行
ny = (nx === 0) ? y + 1 : y;
// 如果已经填充，则进入下一个点
if (board[y][x] !== '.') return fill(board, nx, ny);
// 没有被填充过
for (let i = 1; i <= 9; i++) {
let bx = Math.floor(x / 3),
by = Math.floor(y / 3),
box_key = by * 3 + bx;
if (!rows[y][i] && !cols[x][i] && !boxes[box_key][i]) {
rows[y][i] = 1;
cols[x][i] = 1;
boxes[box_key][i] = 1;
board[y][x] = i.toString();
console.log(board[y][x])
// 递归向下一个点求解
if (fill(board, nx, ny)) return true;
// 恢复初始状态
board[y][x] = '.';
boxes[box_key][i] = 0;
rows[y][i] = 0;
cols[x][i] = 0;
}
}
return false;
}
console.log(board);
};
The problem with fill(), at least with object, is that it passes the same object, by reference, to all element of the array. So if you mutate this object, then it will mutate every object of every arrays.
Note that in your case, you are creating a new Array object using it's constructor ( new Array() ) which makes them objects.
const matrix = new Array(5).fill(new Array(5).fill(0));
console.log(matrix);
In the previous snippet, you can see that the values of the other rows, from the second one to the end, are reference to the initial row.
To get around that, you can fill you array with empty values and then use the map() to create unique object for each position in the array.
const matrix = new Array(5).fill().map(function() { return new Array(5).fill(0); });
console.log(matrix);
As you can see in the previous snippet, all the rows are now their unique reference.
This is the reason all of your values were changed.
I've applied this solution to your code. I wasn't able to test it, because I wasn't sure of the initial parameters to pass.
I've also used anonymous function here ( function() { return; } ), but I would success using arrow function ( () => {} ) instead, if you are comfortable with them. It's cleaner.
var solveSudoku = function (board) {
let rows = new Array(9).fill().map(function() { return new Array(10).fill(0); }),
cols = new Array(9).fill().map(function() { return new Array(10).fill(0); }),
boxes = new Array(9).fill().map(function() { return new Array(10).fill(0); });
for (let i = 0; i < 9; i++) {
for (let j = 0; j < 9; j++) {
let c = board[i][j];
if (c !== '.') {
let n = parseInt(c),
bx = Math.floor(j / 3),
by = Math.floor(i / 3);
// 0代表为使用，1为使用过
rows[i][n] = 1;
console.log(i, n)
cols[j][n] = 1;
// box索引
boxes[by * 3 + bx][n] = 1;
}
}
}
fill(board, 0, 0)
function fill(board, x, y) {
// 完成填充条件
if (y === 9) return true;
// 下一个点的坐标
let nx = (x + 1) % 9,
// 判断进入是否下一行
ny = (nx === 0) ? y + 1 : y;
// 如果已经填充，则进入下一个点
if (board[y][x] !== '.') return fill(board, nx, ny);
// 没有被填充过
for (let i = 1; i <= 9; i++) {
let bx = Math.floor(x / 3),
by = Math.floor(y / 3),
box_key = by * 3 + bx;
if (!rows[y][i] && !cols[x][i] && !boxes[box_key][i]) {
rows[y][i] = 1;
cols[x][i] = 1;
boxes[box_key][i] = 1;
board[y][x] = i.toString();
console.log(board[y][x])
// 递归向下一个点求解
if (fill(board, nx, ny)) return true;
// 恢复初始状态
board[y][x] = '.';
boxes[box_key][i] = 0;
rows[y][i] = 0;
cols[x][i] = 0;
}
}
return false;
}
console.log(board);
};
My code currently has a bug where my 2-d array with the bool value false suddenly contains true values before it is assigned any. My current guesses is either console.log somehow is delayed and picks up the values after it is called, with the updated values or that there is some issue that I don't understand about how scope works in javascript.
As seen below console.log(visited[i][j]) results in false for all values but the
new visited line contains true values even before the following is called.
const field_size = 800;
const cells_in_row = 5;
const frames_per_second = 1;
const cell_size = field_size / cells_in_row;
class Cell {
constructor(x,y) {
this.value = 0;
this.x = x;
this.y = y;
this.coordinates = [x*cell_size,y*cell_size];
}
fill() {
this.value = 1;
}
clear() {
this.value = 0;
}
}
const get_new_grid = (random = 0) => {
const grid = new Array(cells_in_row);
for (let i = 0; i < grid.length; i++) {
grid[i] = new Array(cells_in_row);
for (let j = 0; j < grid.length; j++) {
grid[i][j] = new Cell(i,j);
v = 0;
if (random) {
v = Math.floor(Math.random() * 2);
}
grid[i][j].value = v;
}
}
return grid;
}
const get_islands = (grid) => {
// bool array to mark visited cells
let visited = new Array(cells_in_row);
for (let i = 0; i < grid.length; i++) {
visited[i] = new Array(cells_in_row);
for (let j = 0; j < grid[0].length; j++) {
visited[i][j] = false;
}
}
console.log("New Visited", visited);
let count = 0;
let islands = [];
let island_coords = [];
for (let i = 0; i < grid.length; i++) {
for (let j = 0; j < grid.length; j++) {
if (visited[i][j] == false && grid[i][j].value == 1) {
// visit all cells in this island and increment island count
// dfs will return array of coordinates of island
[visited, island_coords] = dfs(i, j, grid, visited, island_coords);
console.log(visited);
islands.push(island_coords);
count += 1;
}
}
}
return [count, islands];
}
const dfs = (i, j, grid, visited, island_coords) => {
let row_nbr = [-1, -1, -1, 0, 0, 1, 1, 1];
let col_nbr = [-1, 0, 1, -1, 1, -1, 0, 1];
visited[i][j] = true;
island_coords.push([i,j]);
for (let k = 0; k < 8; k++) {
if (is_safe(i + row_nbr[k], j + col_nbr[k], grid, visited)) {
console.log("DFSing " + i + "," + j);
[visited, island_coords] = dfs(i + row_nbr[k], j + col_nbr[k],
grid, visited, island_coords);
}
}
return [visited, island_coords];
}
const is_safe = (i, j, grid, visited) => {
return (i >= 0 && i < grid.length &&
j >= 0 && j < grid.length &&
!(visited[i][j]) && grid[i][j].value === 1);
}
(function () {
var old = console.log;
var logger = document.getElementById('log');
console.log = function () {
for (var i = 0; i < arguments.length; i++) {
if (typeof arguments[i] == 'object') {
logger.innerHTML += (JSON && JSON.stringify ? JSON.stringify(arguments[i], undefined, 2) : arguments[i]) + '<br />';
} else {
logger.innerHTML += arguments[i] + '<br />';
}
}
}
})();
window.onload = () => {
const canvas = document.getElementById('canvas');
const grid = get_new_grid(random = 0);
grid[0][0].value = true;
grid[0][1].value = true;
grid[1][0].value = true;
grid[1][1].value = true;
const islands = get_islands(grid);
console.log(grid);
console.log(islands);
}
<!DOCTYPE html>
<html>
<body>
<script src="gameoflife.js"></script>
<pre id="log"></pre>
</body>
</html>
EDIT:
So I updated the snippet but it looks like it works on this end, however it shows the behavior I mentioned before on my own browser even with the exact same javascript code and html in the snippet.
Mentioned in the comments by Niet, objects logged to the console are live.
So... If I input:
4 1 5 3
INSTEAD OF 1,3,4,5
I GET [ 4, 1, 5, 3 ]
Following is the code for merge sort but for the last comparison the program doesn't fetch updated (1,4) (3,5) value rather (4,1) (5,3) thus giving the wrong result.
var a = [4, 1, 5, 3];
q(a);
function q(a) {
var start = 0;
var n = a.length;
var length = parseInt(n / 2);
if (n < 2) {
return n;
}
var l = [], r = [];
for (i = 0; i < length; i++) {
l[i] = a[i]; //left array
}
for (i = 0, j = length; j < n; i++ , j++) {
r[i] = a[j]; //right array
}
q(l); //merge sort left array
q(r); //merge sort right array
comp(l, r);
}
function comp(l, r) {
var k = [], m = 0, i = 0, j = 0;
while (i < ((l.length)) && j < ((r.length))) {
if (l[i] < r[j]) {
k[m] = l[i];
i++;
m++
}
else {
k[m] = r[j];
j++;
m++
}
}
while (i != (l.length)) {
k[m] = l[i];
m++;
i++;
}
while (j != (r.length)) {
k[m] = r[j];
m++;
j++;
}
console.log(k); //for final output it is [ 4, 1, 5, 3 ] instead of [1,3,4,5]
}
You have a couple small problems. The main one is that you are returning the wrong thing from your edge condition:
if (n < 2) {
return n; // n is just a length; doesn't make sense to return it.
}
n is the length, you really want to return the small array here:
if (n < 2) {
return a; // return the array instead
}
Also, you need to pass the result of the recursive call to your comp function. Right now you're just returning the original lists with:
comp(l, r)
Something like this would work better:
let l_sort = q(l); //merge sort left array
let r_sort = q(r); //merge sort right array
return comp(l_sort, r_sort); // merge the arrays when recursion unwinds.
And you need to return things for recursion to work.
Put all together:
function q(a) {
var start = 0;
var n = a.length;
var length = parseInt(n / 2);
if (n < 2) {
return a;
}
var l = [],
r = [];
for (i = 0; i < length; i++) {
l[i] = a[i]; //left array
}
for (i = 0, j = length; j < n; i++, j++) {
r[i] = a[j]; //right array
}
let l_sort = q(l); //merge sort left array
let r_sort = q(r); //merge sort right array
return comp(l_sort, r_sort);
}
function comp(l, r) {
var k = [],
m = 0,
i = 0,
j = 0;
while (i < ((l.length)) && j < ((r.length))) {
if (l[i] < r[j]) {
k[m] = l[i];
i++;
m++
} else {
k[m] = r[j];
j++;
m++
}
}
while (i != (l.length)) {
k[m] = l[i];
m++;
i++;
}
while (j != (r.length)) {
k[m] = r[j];
m++;
j++;
}
return k
}
console.log(q([4, 1, 5, 3]).join(','));
console.log(q([5, 4, 3, 2, 1]).join(','));
console.log(q([2, 3]).join(','));
console.log(q([3, 2]).join(','));
console.log(q([1]).join(','));
What is a fast and simple implementation of interleave:
console.log( interleave([1,2,3,4,5,6] ,2) ); // [1,4,2,5,3,6]
console.log( interleave([1,2,3,4,5,6,7,8] ,2) ); // [1,5,2,6,3,7,4,8]
console.log( interleave([1,2,3,4,5,6] ,3) ); // [1,3,5,2,4,6]
console.log( interleave([1,2,3,4,5,6,7,8,9],3) ); // [1,4,7,2,5,8,3,6,9]
This mimics taking the array and splitting it into n equal parts, and then shifting items off the front of each partial array in sequence. (n=2 simulates a perfect halving and single shuffle of a deck of cards.)
I don't much care exactly what happens when the number of items in the array is not evenly divisible by n. Reasonable answers might either interleave the leftovers, or even "punt" and throw them all onto the end.
function interleave( deck, step ) {
var copyDeck = deck.slice(),
stop = Math.floor(copyDeck.length/step),
newDeck = [];
for (var i=0; i<step; i++) {
for (var j=0; j<stop; j++) {
newDeck[i + (j*step)] = copyDeck.shift();
}
}
if(copyDeck.length>0) {
newDeck = newDeck.concat(copyDeck);
}
return newDeck;
}
It could be done with a counter instead of shift()
function interleave( deck, step ) {
var len = deck.length,
stop = Math.floor(len/step),
newDeck = [],
cnt=0;
for (var i=0; i<step; i++) {
for (var j=0; j<stop; j++) {
newDeck[i + (j*step)] = deck[cnt++];
}
}
if(cnt<len) {
newDeck = newDeck.concat(deck.slice(cnt,len));
}
return newDeck;
}
And instead of appending the extras to the end, we can use ceil and exit when we run out
function interleave( deck, step ) {
var copyDeck = deck.slice(),
stop = Math.ceil(copyDeck.length/step),
newDeck = [];
for (var i=0; i<step; i++) {
for (var j=0; j<stop && copyDeck.length>0; j++) {
newDeck[i + (j*step)] = copyDeck.shift();
}
}
return newDeck;
}
can i has prize? :-D
function interleave(a, n) {
var i, d = a.length + 1, r = [];
for (i = 0; i < a.length; i++) {
r[i] = a[Math.floor(i * d / n % a.length)];
}
return r;
}
according to my tests r.push(... is faster than r[i] = ... so do with that as you like..
note this only works consistently with sets perfectly divisible by n, here is the most optimized version i can come up with:
function interleave(a, n) {
var i, d = (a.length + 1) / n, r = [a[0]];
for (i = 1; i < a.length; i++) {
r.push(a[Math.floor(i * d) % a.length]);
}
return r;
}
O(n-1), can anyone come up with a log version? to the mathmobile! [spinning mathman logo]
Without for loops (I've added some checkup for the equal blocks):
function interleave(arr, blocks)
{
var len = arr.length / blocks, ret = [], i = 0;
if (len % 1 != 0) return false;
while(arr.length>0)
{
ret.push(arr.splice(i, 1)[0]);
i += (len-1);
if (i>arr.length-1) {i = 0; len--;}
}
return ret;
}
alert(interleave([1,2,3,4,5,6,7,8], 2));
And playground :) http://jsfiddle.net/7tC9F/
how about functional with recursion:
function interleave(a, n) {
function f(a1, d) {
var next = a1.length && f(a1.slice(d), d);
a1.length = Math.min(a1.length, d);
return function(a2) {
if (!a1.length) {
return false;
}
a2.push(a1.shift());
if (next) {
next(a2);
}
return true;
};
}
var r = [], x = f(a, Math.ceil(a.length / n));
while (x(r)) {}
return r;
}
Phrogz was pretty close, but it didn't interleave properly. This is based on that effort:
function interleave(items, parts) {
var len = items.length;
var step = len/parts | 0;
var result = [];
for (var i=0, j; i<step; ++i) {
j = i
while (j < len) {
result.push(items[j]);
j += step;
}
}
return result;
}
interleave([0,1,2,3], 2); // 0,2,1,3
interleave([0,1,2,3,4,5,6,7,8,9,10,11], 2) // 0,6,1,7,2,8,3,9,4,10,5,11
interleave([0,1,2,3,4,5,6,7,8,9,10,11], 3) // 0,4,8,1,5,9,2,6,10,3,7,11
interleave([0,1,2,3,4,5,6,7,8,9,10,11], 4) // 0,3,6,9,1,4,7,10,2,5,8,11
interleave([0,1,2,3,4,5,6,7,8,9,10,11], 5) // 0,2,4,6,8,10,1,3,5,7,9,11
Since I've been pushed to add my own answer early (edited to fix bugs noted by RobG):
function interleave(items,parts){
var stride = Math.ceil( items.length / parts ) || 1;
var result = [], len=items.length;
for (var i=0;i<stride;++i){
for (var j=i;j<len;j+=stride){
result.push(items[j]);
}
}
return result;
}
try this one:
function interleave(deck, base){
var subdecks = [];
for(count = 0; count < base; count++){
subdecks[count] = [];
}
for(var count = 0, subdeck = 0; count < deck.length; count++){
subdecks[subdeck].push(deck[count]);
subdeck = subdeck == base - 1? 0 : subdeck + 1;
}
var newDeck = [];
for(count = 0; count < base; count++){
newDeck = newDeck.concat(subdecks[count]);
}
return newDeck;
}