JavaScript fun: Spiral Matrix

Given an n * n two-dimensional array, use the spiral matrix algorithm to traverse it and return the path. Example: A two-dimensional array of n * n is given. The spiral matrix algorithm is used to traverse the array and return the path.

Example:

array = [[1,2,3],        [4,5,6],        [7,8,9]]  snail(array) // => [1,2,3,6,9,8,7,4,5]

The example of this program may not be intuitive, so the previous figure shows the process.

As you can see, it is like a person moving forward in the maze. It is located at () first, and then goes to the right. when it encounters a boundary, it goes down and encounters a boundary, move to the left ....

Careful people will soon find a rule:

This person in the Maze has four basic travel strategies.

1. When you go to the right, if you encounter a boundary, or the grid on the right has already passed, go down. Otherwise, continue to the right.

2. When you go down, if you encounter a boundary, or the grid below has already passed, you will go to the left. Otherwise, you will continue to go down.

3. When you move to the left, if you encounter a boundary, or the grid on the left has already passed, go up. Otherwise, continue to the left.

4. When you go up, if you encounter a boundary, or the grid above has already passed, you will go to the right. Otherwise, you will continue to go up.

As you can see, this is a recursive process. When will it end?

When this person has accessed each grid, it will be terminated.

The program I wrote below uses a boolean two-dimensional array to record whether the lattice has passed.

Four types of functions are introduced, which indicate four policies.

Function snail il (array) {// current row var row = 0; // current column var col = 0; // the corresponding two-dimensional array that stores boolean values var hasVisited = []; // array for storing results var result = []; // Number of array elements var size = array. length * array [0]. length; // boolean two-dimensional array initialization for (var I = 0; I <array. length; I ++) {var temp = []; var len = array [I]. length; for (var j = 0; j <len; j ++) {temp. push (false);} hasVisited. push (temp);} function right () {if (result. length <size) {result. push (array [row] [col]); hasVisited [row] [col] = true; if (col + 1> = array. length | hasVisited [row] [col + 1]) {row + = 1; down ();} else {col + = 1; right ();}}} function down () {if (result. length <size) {result. push (array [row] [col]); hasVisited [row] [col] = true; if (row + 1> = array. length | hasVisited [row + 1] [col]) {col-= 1; left () ;}else {row + = 1; down ();}}} function left () {if (result. length <size) {result. push (array [row] [col]); hasVisited [row] [col] = true; if (col = 0 | hasVisited [row] [col-1]) {row-= 1; up () ;}else {col-= 1; left () ;}} function up () {if (result. length <size) {result. push (array [row] [col]); hasVisited [row] [col] = true; if (hasVisited [row-1] [col]) {col + = 1; right () ;}else {row-= 1; up () ;}}// first right (); return result ;}

The above is the JavaScript interesting question: the content of the spiral matrix. For more information, see the PHP Chinese website (www.php1.cn )!