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[LeetCode]566. Reshape the Matri

[LeetCode]566. Reshape the Matri

作者: Eazow | 来源:发表于2017-05-10 12:21 被阅读18次
题目

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.
难度

Easy

方法

转换后的矩阵和之前的矩阵不一样大时,返回之前的矩阵。否则将之前矩阵的数据依次填入新的矩阵中

python代码
class Solution(object):
    def matrixReshape(self, nums, r, c):
        """
        :type nums: List[List[int]]
        :type r: int
        :type c: int
        :rtype: List[List[int]]
        """
        numsRows = len(nums)
        numsCols = len(nums[0])
        if numsRows*numsCols != r*c:
            return nums
        result = [[0 for col in range(c)] for row in xrange(r)]
        i = 0
        while i < r*c:
            result[i/c][i%c] = nums[i/numsCols][i%numsCols]
            i += 1

        return result

assert Solution().matrixReshape([[1,2],[3,4]], 1, 4) == [[1,2,3,4]]
assert Solution().matrixReshape([[1,2],[3,4]], 2, 4) == [[1,2],[3,4]]
assert Solution().matrixReshape([[1,2,3,4]], 2, 2) == [[1,2],[3,4]]

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