Array-Reshape the Matrix

最新推荐文章于 2022-10-27 20:34:54 发布

kevin聪

最新推荐文章于 2022-10-27 20:34:54 发布

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分类专栏：算法与数据结构 leetcode 文章标签：算法数据结构 leetcode

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介绍了一种在保持原始数据的前提下，将矩阵重塑为不同尺寸的新矩阵的方法。提供了三种解决方案，包括将二维数组转换为一维数组再重塑的方法，以及使用行和列指针进行重塑的方法。

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Problem Description:
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.

My Solution:

Time Complexity:O(m*n)
Space Complexity:O(m*n),temp is of size m*n
class Solution {
    public int[][] matrixReshape(int[][] nums, int r, int c) {
        int rowN = nums.length;
        int columnN = nums[0].length;
        if((r * c) != (rowN * columnN)){
            return nums;                            
        }
        int[] temp  = new int[rowN * columnN];
        int sum = 0;
        for(int i = 0;i < rowN;i++){
            for(int j = 0;j < columnN;j++){
                temp[sum++] = nums[i][j];
            }
        }
        int[][] arr = new int[r][c];
        for(int i = 0;i < r;i++){
            for(int j = 0;j < c;j++){
                arr[i][j] = temp[i*c + j];
            }
        }
        return arr;
    }
}

Some Better Solutions:
1

Time Complexity:O(m*n)
Space Complexity:O(m*n),res is of size m*n
public class Solution {
    public int[][] matrixReshape(int[][] nums, int r, int c) {
        int[][] res = new int[r][c];
        if (nums.length == 0 || r * c != nums.length * nums[0].length)
            return nums;
        int rows = 0, cols = 0;
        for (int i = 0; i < nums.length; i++) {
            for (int j = 0; j < nums[0].length; j++) {
                res[rows][cols] = nums[i][j];
                cols++;
                if (cols == c) {
                    rows++;
                    cols = 0;
                }
            }
        }
        return res;
    }
}

Time Complexity:O(m*n)
Space Complexity:O(m*n)
public class Solution {
    public int[][] matrixReshape(int[][] nums, int r, int c) {
        int[][] res = new int[r][c];
        if (nums.length == 0 || r * c != nums.length * nums[0].length)
            return nums;
        int count = 0;
        for (int i = 0; i < nums.length; i++) {
            for (int j = 0; j < nums[0].length; j++) {
                res[count / c][count % c] = nums[i][j];
                count++;
            }
        }
        return res;
    }
}