LeetCode 566 Reshape the Matrix

最新推荐文章于 2025-09-12 17:31:31 发布

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原文链接：http://www.cnblogs.com/hzg1981/p/8818298.html

文章标签：

#数据结构与算法 #matlab

本文介绍了一个在给定条件下重塑矩阵的算法。该算法首先检查重塑是否合法，然后将原矩阵的数据按行顺序转换为一维数组，最后根据目标形状重新构造矩阵。文章提供了两个示例并展示了两种不同的实现方式。

题目：

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.

Note:

1. The height and width of the given matrix is in range [1, 100].
2. The given r and c are all positive.

分析：

提交：

class Solution {
    public int[][] matrixReshape(int[][] nums, int r, int c) {
        int m=nums.length;
        int n=nums[0].length;
        
        if(r*c == m*n) {
            
            
            int[] array = new int[m*n];
            for(int i=0; i<m; i++) {
                for(int j=0; j<n; j++) {
                    array[i*n+j] = nums[i][j];
                }
            }
            
            int[][] newMatrix = new int[r][c];
            for(int i=0; i<r; i++){
                for(int j=0; j<c; j++) {
                    newMatrix[i][j] = array[i*c+j];
                }
            }
            
            return newMatrix;
        } else {
            return nums;
        }
    }
}

结果：

略作修改，二次提交：

class Solution {
    public int[][] matrixReshape(int[][] nums, int r, int c) {
        int m=nums.length;
        int n=nums[0].length;
        
        if(r*c != m*n) {
            return nums;
        }
            
       int[] array = new int[m*n];
       for(int i=0; i<m; i++) {
           for(int j=0; j<n; j++) {
               array[i*n+j] = nums[i][j];
           }
       }
       
       int[][] newMatrix = new int[r][c];
       for(int i=0; i<r; i++){
           for(int j=0; j<c; j++) {
               newMatrix[i][j] = array[i*c+j];
           }
       }
       
       
      /* int[][] newMatrix = new int[r][c];
       for(int i=0; i<r; i++){
           for(int j=0; j<c; j++) {
               newMatrix[i][j] = nums[i]*c+j];
           }
       }
        */
        
       return newMatrix;


            
   
    }
}