当前位置 : 首页 » 文章分类 :  算法  »  LeetCode.566.Reshape the Matrix

LeetCode.566.Reshape the Matrix


题目描述

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.

解题过程

很简单的二维数组操作题

package _566_ReshapeTheMatrix;

class Solution {
    public int[][] matrixReshape(int[][] nums, int r, int c) {
        int origin_row = nums.length;
        int origin_column = nums[0].length;
        if(origin_row*origin_column != r*c){
            return nums;
        }
        int[][] output = new int[r][];
        output[0] = new int[c];
        int m=0, n=0;
        for(int i=0; i<origin_row; i++)
            for(int j=0; j<origin_column; j++){
                if(n>=c){//新矩阵换行
                    n=0;
                    m++;
                    output[m] = new int[c];
                }
                output[m][n++] = nums[i][j];
            }
        return output;
    }
}

public class ReshapeTheMatrix {
    public static void main(String[] args){
        Solution solution = new Solution();
        int[][] input = {{1,2},{3,4}};
        int[][] output = solution.matrixReshape(input, 4, 1);
        for(int i=0; i<output.length; i++){
            for(int j=0; j<output[0].length; j++){
                System.out.print(output[i][j]+" ");
            }
            System.out.println();
        }            
    }
}

提交后看别人代码,人家都是int[][] arr = new int[r][c];直接就申请好新二维数组的空间了,我是每到一个新的行再申请这行的空间。

关于新数组中下标的获取,看到一个用当前遍历个数对行做除、对列取模的方法,很巧妙:

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;
    }
}

GitHub代码


上一篇 LeetCode.237.Delete Node in a Linked List

下一篇 LeetCode.575.Distribute Candies

阅读
542
阅读预计3分钟
创建日期 2018-01-18
修改日期 2018-01-18
类别
百度推荐