Algorithm
Problem Name: 566. Reshape the Matrix
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
Code Examples
#1 Code Example with C++ Programming
Code -
C++ Programming
start coding..
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));
int row = 0, col = 0;
for(int i = 0; i < r; i++)
for(int j = 0; j < c; j++){
res[i][j] = nums[row][col++];
if(col == n> col = 0, row++;
}
return res;
}
};
Input
Output
#2 Code Example with Java Programming
Code -
Java Programming
class Solution {
public int[][] matrixReshape(int[][] nums, int r, int c) {
int numOfElem = nums.length*nums[0].length;
if (numOfElem != r*c) {
return nums;
}
int[] expandedArray = new int[numOfElem];
int k = 0;
for (int i=0;i < nums.length;i++) {
for (int j=0;j < nums[i].length;j++) {
expandedArray[k] = nums[i][j];
k++;
}
}
int[][] ans = new int[r][c];
int i=0;
int j=0;
for(k=0;k < numOfElem;k++) {
ans[i][j] = expandedArray[k];
j++;
if (j == c) {
j = 0;
i++;
}
}
return ans;
}
}
Input
Output
#3 Code Example with Javascript Programming
Code -
Javascript Programming
const matrixReshape = function(nums, r, c) {
if (isValid(nums, r, c) === false) {
return nums;
}
const arr = [];
nums.forEach(el => arr.push(...el));
const res = [];
for (let start = 0; start < arr.length; start = start + c) {
res.push(arr.slice(start, start + c));
}
return res;
};
function isValid(matrix, r, c) {
if (matrix.length * matrix[0].length !== r * c) {
return false;
} else {
return true;
}
}
console.log(matrixReshape([[1, 2], [3, 4]], 1, 4));
console.log(matrixReshape([[1, 2], [3, 4]], 2, 4));
Input
Output
#4 Code Example with Python Programming
Code -
Python Programming
start coding...
class Solution:
def matrixReshape(self, nums, r, c):
"""
:type nums: List[List[int]]
:type r: int
:type c: int
:rtype: List[List[int]]
"""
nums_ordered=[x for y in nums for x in y]
if r*c==len(nums)*len(nums[0]):
return [nums_ordered[c*i:c*(i+1)] for i in range(r)]
else:return nums
Input
Output
#5 Code Example with C# Programming
Code -
C# Programming
using System.Linq;
namespace LeetCode
{
public class _0567_PermutationInString
{
public bool CheckInclusion(string s1, string s2)
{
var s1Length = s1.Length;
var s2Length = s2.Length;
if (s1Length > s2Length) return false;
var s1Count = new int[26];
foreach (var ch in s1)
s1Count[ch - 'a']++;
var s2Count = new int[26];
for (int i = 0; i < s2Length; i++)
{
s2Count[s2[i] - 'a']++;
if (i >= s1Length)
s2Count[s2[i - s1Length] - 'a']--;
if (s2Count.SequenceEqual(s1Count))
return true;
}
return false;
}
}
}
Input
Output