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SLPHJava：public class Solution { public int[][] matrixReshape(int[][] nums, int r, int c) { int original_r = nums.length; int original_c = nums[0].length; if (original_r * original_c == r * c) { int[][] result = new int[r][c]; for (int i = 0; i < r * c; i++) { result[i / c][i % c] = nums[i / original_c][i % original_c]; } return result; } else { return nums; } } } -
SLPHC++解法一：
class Solution { public: vector<vector<int>> matrixReshape(vector<vector<int>>& nums, int r, int c) { int m = nums.size(), n = nums[0].size(); if (m * n != r * c) return nums; vector<vector<int>> res(r, vector<int>(c)); for (int i = 0; i < r; ++i) { for (int j = 0; j < c; ++j) { int k = i * c + j; res[i][j] = nums[k / n][k % n]; } } return res; } }; -
SLPHC++解法二：
class Solution { public: vector<vector<int>> matrixReshape(vector<vector<int>>& nums, int r, int c) { int m = nums.size(), n = nums[0].size(); if (m * n != r * c) return nums; vector<vector<int>> res(r, vector<int>(c)); for (int i = 0; i < r * c; ++i) { res[i / c][i % c] = nums[i / n][i % n]; } return res; } }; -
原题： https://leetcode.com/problems/reshape-the-matrix/description/
题意： 给定二维矩阵nums，将其转化为r行c列的新矩阵。若无法完成转化，返回原矩阵。
约定：（1）给定矩阵的高度和宽度范围[1, 100]；（2）r和c都是正数。
例子：
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.
标签： matrix、nums、reshape、矩阵、row、面试
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