reshape-the-matrix.py

# Time:  O(m * n)
# Space: O(m * n)

# In MATLAB, there is a very useful function called 'reshape',
# which can reshape a matrix into a new one with different size but keep its original data.
#
# You're given a matrix represented by a two-dimensional array,
# and two positive integers r and c representing the row number
# and column number of the wanted reshaped matrix, respectively.
#
# The reshaped matrix need to be filled with
# all the elements of the original matrix in the same row-traversing order as they were.
#
# If the 'reshape' operation with given parameters is possible and legal,
# output the new reshaped matrix; Otherwise, output the original matrix.
#
# Example 1:
# Input:
# nums =
# [[1,2],
#  [3,4]]
# r = 1, c = 4
# Output:
# [[1,2,3,4]]
# Explanation:
# The row-traversing of nums is [1,2,3,4]. The new reshaped matrix is a 1 * 4 matrix,
# fill it row by row by using the previous list.
#
# Example 2:
# Input:
# nums =
# [[1,2],
#  [3,4]]
# r = 2, c = 4
# Output:
# [[1,2],
#  [3,4]]
# Explanation:
# There is no way to reshape a 2 * 2 matrix to a 2 * 4 matrix. So output the original matrix.
#
# Note:
# The height and width of the given matrix is in range [1, 100].
# The given r and c are all positive.

class Solution(object):
    def matrixReshape(self, nums, r, c):
        """
        :type nums: List[List[int]]
        :type r: int
        :type c: int
        :rtype: List[List[int]]
        """
        if not nums or \
           r*c != len(nums) * len(nums[0]):
            return nums

        result = [[0 for _ in xrange(c)] for _ in xrange(r)]
        count = 0
        for i in xrange(len(nums)):
            for j in xrange(len(nums[0])):
                result[count/c][count%c] = nums[i][j]
                count += 1
        return result