【CodeWars】Snail Sort

地址：https://www.codewars.com/kata/521c2db8ddc89b9b7a0000c1/train/javascript
Snail Sort
Given an n x n array, return the array elements arranged from outermost elements to the middle element, traveling clockwise.

array = [[1,2,3],
[4,5,6],
[7,8,9]]
snail(array) #=> [1,2,3,6,9,8,7,4,5]
For better understanding, please follow the numbers of the next array consecutively:

array = [[1,2,3],
[8,9,4],
[7,6,5]]
snail(array) #=> [1,2,3,4,5,6,7,8,9]
This image will illustrate things more clearly:

NOTE: The idea is not sort the elements from the lowest value to the highest; the idea is to traverse the 2-d array in a clockwise snailshell pattern.

NOTE 2: The 0x0 (empty matrix) is represented as en empty array inside an array [[]].

我的答案：

// 规律：（0,0）（0,1）j变化，递增
//      （0,2）（1,2）i变化，递增
//      （2,2）（2,1）j变化，递减
const snail = (array) => {
	const res = [];
	let len = array.length;
	if (len === 1) {
		return array[0];
	}
	let times = 0;
	let i = 0;
	let j = 0;
	let k = 0;
	for (times = len; times > 0; times = times - 2) {
		// 最外层循环，走完一次就是外层的一圈
		// i,j初始值由(0,0)变成(1,1)，打印一圈，ij的初始位置就往里挪动一位
		for (k = 1; k < times; k++, j++) {
			res.push(array[i][j]);
		}
		for (k = 1; k < times; k++, i++) {
			res.push(array[i][j]);
		}
		for (k = 1; k < times; k++, j--) {
			res.push(array[i][j]);
		}
		for (k = 1; k < times; k++, i--) {
			res.push(array[i][j]);
		}
		// 为了让下一个循环的ij位置从(1,1)开始，分别加1
		i = i + 1;
		j = j + 1;
	}
	if (len % 2 === 1) {
		i = i - 1;
		j = j - 1;
		res.push(array[i][j]);
	}
	return res;
};
console.log(
	snail([
		[1, 2, 3],
		[4, 5, 6],
		[7, 8, 9],
	])
);

打印结果：[
  1, 2, 3, 6, 9,
  8, 7, 4, 5
]