135 - 54 螺旋矩阵

题目

给定一个包含 m x n 个元素的矩阵（m 行, n 列），请按照顺时针螺旋顺序，返回矩阵中的所有元素。

示例 1:

输入: [ [ 1, 2, 3 ], [ 4, 5, 6 ], [ 7, 8, 9 ] ] 输出: [1,2,3,6,9,8,7,4,5]

示例 2:

输入: [ [1, 2, 3, 4], [5, 6, 7, 8], [9,10,11,12] ] 输出: [1,2,3,4,8,12,11,10,9,5,6,7]

解答

https://leetcode-cn.com/problems/spiral-matrix/solution/na-yi-xing-ni-shi-zhen-zhuan-yi-xia-by-suixin-3/

感觉python有很多函数，专门为刷题发明的😂😂

class Solution:
    def spiralOrder(self, matrix: List[List[int]]) -> List[int]:
        res = []
        while matrix:
            res += matrix.pop(0)
            matrix = list(map(list, zip(*matrix)))[::-1]
        return res

Runtime: 20 ms, faster than 100.00% of Python3 online submissions for Spiral Matrix.

Memory Usage: 12.8 MB, less than 100.00% of Python3 online submissions for Spiral Matrix.

关键是zip能把两个函数竖着拼起来，然后逆序一下再输出就行了

方法还是绕着外面一点点算进去。。

https://leetcode-cn.com/problems/spiral-matrix/solution/luo-xuan-ju-zhen-by-liao-tian-yi-jian/

var spiralOrder = function(matrix) {
  if (matrix.length === 0) {
    return []
  }
  const ans = []
  let u = 0,
    d = matrix.length - 1,
    l = 0,
    r = matrix[0].length - 1
  while (1) {
    for (let i = l; i <= r; ++i) {
      ans.push(matrix[u][i])
    }
    if (++u > d) {
      break
    }
    for (let i = u; i <= d; ++i) {
      ans.push(matrix[i][r])
    }
    if (--r < l) {
      break
    }
    for (let i = r; i >= l; --i) {
      ans.push(matrix[d][i])
    }
    if (--d < u) {
      break
    }
    for (let i = d; i >= u; --i) {
      ans.push(matrix[i][l])
    }
    if (++l > r) {
      break
    }
  }
  return ans
};

Runtime: 48 ms, faster than 88.01% of JavaScript online submissions for Spiral Matrix.

Memory Usage: 33.8 MB, less than 36.36% of JavaScript online submissions for Spiral Matrix.

var spiralOrder = function(matrix) {
  if (matrix.length === 0) {
    return []
  }
  const list = []
  let m = matrix.length
  n = matrix[0].length
  i = 0
  count = ~~((Math.min(m, n) + 1) / 2)
  while (i < count) {
    for (let j = i; j < n - i; j++) {
      list.push(matrix[i][j]);
    }
    for (let j = i + 1; j < m - i; j++) {
      list.push(matrix[j][(n - 1) - i]);
    }

    for (let j = (n - 1) - (i + 1); j >= i && (m - 1 - i != i); j--) {
      list.push(matrix[(m - 1) - i][j]);
    }
    for (let j = (m - 1) - (i + 1); j >= i + 1 && (n - 1 - i) != i; j--) {
      list.push(matrix[j][i]);
    }
    i++;
  }
  return list
};

Runtime: 52 ms, faster than 68.49% of JavaScript online submissions for Spiral Matrix.

Memory Usage: 33.8 MB, less than 63.64% of JavaScript online submissions for Spiral Matrix.