109 - 566重塑矩阵

题目

在MATLAB中，有[一个非常有用的函数 reshape，它可以将一个矩阵重塑为另一个大小不同的新矩阵，但保留其原始数据。](一个非常有用的函数 reshape，它可以将一个矩阵重塑为另一个大小不同的新矩阵，但保留其原始数据。

给出一个由二维数组表示的矩阵，以及两个正整数r和c，分别表示想要的重)

[给出一个由二维数组表示的矩阵，以及两个正整数r和c，分别表示想要的重](一个非常有用的函数 reshape，它可以将一个矩阵重塑为另一个大小不同的新矩阵，但保留其原始数据。

给出一个由二维数组表示的矩阵，以及两个正整数r和c，分别表示想要的重)构的矩阵的行数和列数。

重构后的矩阵需要将原始矩阵的所有元素以相同的行遍历顺序填充。

如果具有给定参数的reshape操作是可行且合理的，则输出新的重塑矩阵；否则，输出原始矩阵。

示例 1:

输入: nums = [[1,2], [3,4]] r = 1, c = 4 输出: [[1,2,3,4]] 解释: 行遍历nums的结果是 [1,2,3,4]。新的矩阵是 1 * 4 矩阵, 用之前的元素值一行一行填充新矩阵。

示例 2:

输入: nums = [[1,2], [3,4]] r = 2, c = 4 输出: [[1,2], [3,4]] 解释: 没有办法将 2 2 矩阵转化为 2 4 矩阵。 所以输出原矩阵。

注意：

• 给定矩阵的宽和高范围在 [1, 100]。

• 给定的 r 和 c 都是正数。

解答

只能想到，全部坍缩成一维数组，然后一个个插进去。。

https://leetcode.com/problems/reshape-the-matrix/discuss/231989/simple-100-js-solution

var matrixReshape = function (nums, r, c) {
  let data = [].concat(...nums)
  if (data.length != r * c) {
    return nums
  }
  let row = []
  let ans = []
  data.forEach(item => {
    row.push(item)
    if (row.length === c) {
      ans.push(row)
      row = []
    }
  });
  return ans
};

Runtime: 76 ms, faster than 91.14% of JavaScript online submissions for Reshape the Matrix.

Memory Usage: 40.2 MB, less than 60.00% of JavaScript online submissions for Reshape the Matrix.

没想到还能有这种[].concat(...nums)这个做法。学到了。。

https://leetcode.com/problems/reshape-the-matrix/discuss/102532/A-few-JavaScript-solutions

js的一行代码版本：

var matrixReshape = function(nums, r, c) {
  let data = [].concat(...nums)
  return data.length === r * c ? new Array(r).fill(0).map((row, r) => data.slice(r * c, r * c + c)) : nums
};

Runtime: 80 ms, faster than 78.16% of JavaScript online submissions for Reshape the Matrix. Memory Usage: 38.9 MB, less than 100.00% of JavaScript online submissions for Reshape the Matrix.

go的二维数组好麻烦。。[][]int{{1, 2}, {3, 4}}要这么定义。。

https://leetcode.com/problems/reshape-the-matrix/discuss/317083/golang-solution

func matrixReshape(nums [][]int, r int, c int) [][]int {
    if len(nums)*len(nums[0]) != c*r {
        return nums
    }
    result := make([][]int, 0)
    for i := 0; i < r; i++ {
        result = append(result, make([]int, c))
    }
    for i := 0; i < c*r; i++ {
        result[i/c][i%c] = nums[i/len(nums[0])][i%len(nums[0])]
    }
    return result
}

Runtime: 60 ms, faster than 52.05% of Go online submissions for Reshape the Matrix.

Memory Usage: 75 MB, less than 100.00% of Go online submissions for Reshape the Matrix.

https://leetcode.com/problems/reshape-the-matrix/discuss/102499/Python-Simple-with-Explanation

class Solution:
    def matrixReshape(self, nums: List[List[int]], r: int, c: int) -> List[List[int]]:
        if len(nums) * len(nums[0]) != r * c:
            return nums
        data = [val for row in nums for val in row]
        ans = [[0] * c for _ in range(r)]
        count = 0
        for i in range(r):
            for j in range(c):
                ans[i][j] = data[count]
                count += 1
        return ans

Runtime: 120 ms, faster than 52.08% of Python3 online submissions for Reshape the Matrix.

Memory Usage: 14.8 MB, less than 5.55% of Python3 online submissions for Reshape the Matrix.

https://leetcode-cn.com/problems/reshape-the-matrix/solution/python-die-dai-qi-by-mao1112/

class Solution:
    def matrixReshape(self, nums: List[List[int]], r: int, c: int) -> List[List[int]]:
        def Y(M):
            for R in M:
                yield from R
        if r*c != len(nums) * len(nums[0]):
            return nums
        it = Y(nums)
        return [[next(it) for _ in range(c)] for _ in range(r)]

Runtime: 172 ms, faster than 5.54% of Python3 online submissions for Reshape the Matrix.

Memory Usage: 14.9 MB, less than 5.55% of Python3 online submissions for Reshape the Matrix.

虽然迭代器看起来很酷，其实很耗成本😂