Description
In MATLAB, there is a handy function called reshape which can reshape an m x n matrix into a new one with a different size r x c keeping its original data.
You are given an m x n matrix mat and two integers r and c representing the number of rows and the number of columns of the wanted reshaped matrix.
The reshaped matrix should 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: mat = [[1,2],[3,4]], r = 1, c = 4
Output: [[1,2,3,4]]
Example 2:
Input: mat = [[1,2],[3,4]], r = 2, c = 4
Output: [[1,2],[3,4]]
Constraints:
m == mat.lengthn == mat[i].length1 <= m, n <= 100-1000 <= mat[i][j] <= 10001 <= r, c <= 300
Solutions
Solution 1: Simulation
First, we get the number of rows and columns of the original matrix, denoted as m and n respectively. If m × n ≠ r × c, then the matrix cannot be reshaped, and we return the original matrix directly.
Otherwise, we create a new matrix with r rows and c columns. Starting from the first element of the original matrix, we traverse all elements in row-major order and place the traversed elements into the new matrix in order.
After traversing all elements of the original matrix, we get the answer.
The time complexity is O(m × n), where m and n are the number of rows and columns of the original matrix, respectively. Ignoring the space consumption of the answer, the space complexity is O(1).
PythonJavaC++GoTypeScriptRustC
class Solution:
def matrixReshape(self, mat: List[List[int]], r: int, c: int) -> List[List[int]]:
m, n = len(mat), len(mat[0])
if m * n != r * c:
return mat
ans = [[0] * c for _ in range(r)]
for i in range(m * n):
ans[i // c][i % c] = mat[i // n][i % n]
return ans(code-box)
class Solution {
public int[][] matrixReshape(int[][] mat, int r, int c) {
int m = mat.length, n = mat[0].length;
if (m * n != r * c) {
return mat;
}
int[][] ans = new int[r][c];
for (int i = 0; i < m * n; ++i) {
ans[i / c][i % c] = mat[i / n][i % n];
}
return ans;
}
}(code-box)
class Solution {
public:
vector<vector<int>> matrixReshape(vector<vector<int>>& mat, int r, int c) {
int m = mat.size(), n = mat[0].size();
if (m * n != r * c) {
return mat;
}
vector<vector<int>> ans(r, vector<int>(c));
for (int i = 0; i < m * n; ++i) {
ans[i / c][i % c] = mat[i / n][i % n];
}
return ans;
}
};(code-box)
func matrixReshape(mat [][]int, r int, c int) [][]int {
m, n := len(mat), len(mat[0])
if m*n != r*c {
return mat
}
ans := make([][]int, r)
for i := range ans {
ans[i] = make([]int, c)
}
for i := 0; i < m*n; i++ {
ans[i/c][i%c] = mat[i/n][i%n]
}
return ans
}(code-box)
function matrixReshape(mat: number[][], r: number, c: number): number[][] {
let m = mat.length,
n = mat[0].length;
if (m * n != r * c) return mat;
let ans = Array.from({ length: r }, v => new Array(c).fill(0));
let k = 0;
for (let i = 0; i < m; ++i) {
for (let j = 0; j < n; ++j) {
ans[Math.floor(k / c)][k % c] = mat[i][j];
++k;
}
}
return ans;
}(code-box)
impl Solution {
pub fn matrix_reshape(mat: Vec<Vec<i32>>, r: i32, c: i32) -> Vec<Vec<i32>> {
let r = r as usize;
let c = c as usize;
let m = mat.len();
let n = mat[0].len();
if m * n != r * c {
return mat;
}
let mut i = 0;
let mut j = 0;
(0..r)
.into_iter()
.map(|_| {
(0..c)
.into_iter()
.map(|_| {
let res = mat[i][j];
j += 1;
if j == n {
j = 0;
i += 1;
}
res
})
.collect()
})
.collect()
}
}(code-box)
/**
* Return an array of arrays of size *returnSize.
* The sizes of the arrays are returned as *returnColumnSizes array.
* Note: Both returned array and *columnSizes array must be malloced, assume caller calls free().
*/
int** matrixReshape(int** mat, int matSize, int* matColSize, int r, int c, int* returnSize, int** returnColumnSizes) {
if (matSize * matColSize[0] != r * c) {
*returnSize = matSize;
*returnColumnSizes = matColSize;
return mat;
}
*returnSize = r;
*returnColumnSizes = malloc(sizeof(int) * r);
int** ans = malloc(sizeof(int*) * r);
for (int i = 0; i < r; i++) {
(*returnColumnSizes)[i] = c;
ans[i] = malloc(sizeof(int) * c);
}
for (int i = 0; i < r * c; i++) {
ans[i / c][i % c] = mat[i / matColSize[0]][i % matColSize[0]];
}
return ans;
}(code-box)
Solution 2
TypeScript
function matrixReshape(mat: number[][], r: number, c: number): number[][] {
const m = mat.length;
const n = mat[0].length;
if (m * n !== r * c) {
return mat;
}
const ans = Array.from({ length: r }, () => new Array(c).fill(0));
for (let i = 0; i < r * c; i++) {
ans[Math.floor(i / c)][i % c] = mat[Math.floor(i / n)][i % n];
}
return ans;
}(code-box)
