leetCode-Reshape the Matrix

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 = 4Output: [[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 = 4Output: [[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*nclass 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*npublic 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;    }}

2

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;    }}

总结:
我的方法相当于暴力枚举,既多了遍历nums将它转换为一维数组的时间，也多了temp的存储空间
方法1多了两个”指针”ros和cols，cols每增加c个,那么置0
方法2与方法一类似,只是用了count，行标相当于count/c,列标相当于count%c