给定一个m x n维的二维数组,我如何以逆时针的方式循环它们?
例如:
matrix = [
[ 1, 2, 3, 4 ],
[ 5, 6, 7, 8 ],
[ 9, 10, 11, 12 ],
[ 13, 14, 15, 16 ]
]
1st loop: 1, 5, 9, 13, 14, 15, 16, 12, 8, 4, 3, 2
2nd loop: 6, 10, 11, 7, 6
我真的不介意这个实现是用ruby还是js给出的
我现在的解决方案是这样的:
(1..rotate).each do
bottom_left_corner = i
top_right_corner = j
start_nth_layer = 1; end_nth_layer = matrix.length - 2
matrix.reverse_each do
matrix[bottom_left_corner].unshift(matrix[bottom_left_corner - 1].shift) if bottom_left_corner > 0
matrix[top_right_corner] << matrix[top_right_corner + 1].pop if top_right_corner < matrix.length - 1
bottom_left_corner -= 1; top_right_corner += 1
end
nth_layer(matrix, start_nth_layer, end_nth_layer)
end
更新
只要输出的顺序正确,输出的格式就无关紧要。
问题的目的
这个问题的目的是逆时针逐层遍历这些数组,直到没有更多的层对于每次遍历,我们都会逆时针移动值。例如:
Iteration 1: Iteration 2: Iteration 3:
============== ============= ==============
1 2 3 4 2 3 4 8 3 4 8 12
5 6 7 8 1 7 11 12 2 11 10 16
9 10 11 12 => 5 6 10 16 => 1 7 6 15
13 14 15 16 9 13 14 15 5 9 13 14
最佳答案
这是矩阵层旋转问题以下是我的完整解决方案:
function matrixRotation(matrix, rotate) {
var r = matrix.length, c = matrix[0].length;
var depth = Math.min(r,c)/2;
for(var i=0;i<depth;i++) {
var x = i, y = i, n = (r-i*2 - 2)*2 + (c-i*2-2)*2+4, dir = 'down', index=0;
var buf = [];
while(index < n) {
buf.push(matrix[x][y]);
if(dir == 'down') {
if(x+1 >= r-i) {
dir = 'right';
y++;
} else {
x++;
}
} else if(dir == 'right') {
if(y+1 >= c-i) {
dir = 'up';
x--;
} else {
y++;
}
} else if(dir == 'up') {
if(x-1 <= i-1) {
dir = 'left';
y--;
} else {
x--;
}
} else if(dir == 'left') {
y--;
}
index++;
}
var move = rotate%n;
buf = [...buf.slice(buf.length-move), ...buf.slice(0, buf.length-move)]
x = i, y = i, dir = 'down', index = 0;
while(index < n) {
matrix[x][y] = buf[index];
if(dir == 'down') {
if(x+1 >= r-i) {
dir = 'right';
y++;
} else {
x++;
}
} else if(dir == 'right') {
if(y+1 >= c-i) {
dir = 'up';
x--;
} else {
y++;
}
} else if(dir == 'up') {
if(x-1 <= i-1) {
dir = 'left';
y--;
} else {
x--;
}
} else if(dir == 'left') {
y--;
}
index++;
}
}
matrix.map(row => console.log(row.join(' ')));
}
const matrix = [[1, 2, 3, 4], [5, 6, 7, 8], [9, 10, 11, 12], [13, 14, 15, 16]]
// rotate count
const r = 3
matrixRotation(matrix, r);