Reshape the Matrix LT566

佚名 6年前 (2019-05-11) 随笔 1174人围观抢沙发百度已收录

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.

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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.

Idea 1. Observe and find the 1-D index is: rowIndex * N + columnIndex,

Time complexity: O(M*N)

Space complexity: O(1)

 1 class Solution {
 2     public int[][] matrixReshape(int[][] nums, int r, int c) {
 3         int M = nums.length;
 4         int N = nums[0].length;
 5         
 6         if(M*N != r*c || M == r) {
 7             return nums;
 8         }
 9         
10         int[][] result = new int[r][c];
11         for(int i = 0; i < M; ++i) {
12             for(int j = 0; j < N; ++j) {
13                 int x = (i * N + j)/c;
14                 int y = (i * N + j)%c;
15                 result[x][y] = nums[i][j];
16             }
17         }
18         
19         return result;
20     }
21 }

Idea 1.b one-loop instead, more conciser code

 1 class Solution {
 2     public int[][] matrixReshape(int[][] nums, int r, int c) {
 3         int M = nums.length;
 4         int N = nums[0].length;
 5         
 6         if(M*N != r*c || M == r) {
 7             return nums;
 8         }
 9         
10         int[][] result = new int[r][c];
11         for(int i = 0; i < M * N; ++i) {
12             result[i/c][i%c] = nums[i/N][i%N];
13         }
14         
15         return result;
16     }
17 }