【LeetCode】566. Reshape the Matrix（重塑矩阵）

【LeetCode】566. Reshape the Matrix（重塑矩阵）

题目链接：https://leetcode.com/problems/reshape-the-matrix/description/

问题描述：直接看例子

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.

解释：给一个矩阵，要求变化成给定的r行c列的矩阵，并输出；否则，输出原矩阵。

例如给定矩阵是3x4，那么可以变化成1x12，12x1，2x6，6x2，4x3，当然也可以和原矩阵相等3x4.总之要求元素个数必须相同。

我的代码：

class Solution {
    public int[][] matrixReshape(int[][] nums, int r, int c) {
        int n=nums.length;
        int m=nums[0].length;
        int arr[][]=new int[r][c];
        if(m*n==r*c)
        {
            int[]num=new int[n*m];
            for(int i=0,k=0;i

标准答案：

public class Solution {  
    public int[][] matrixReshape(int[][] nums, int r, int c) {  
        int original_r = nums.length;  
        int original_c = nums[0].length;  
        if (original_r * original_c == r * c) {  
            int[][] result = new int[r][c];  
            for (int i = 0; i < r * c; i++) {  
                result[i / c][i % c] = nums[i / original_c][i % original_c];  
            }  
            return result;  
        } else {  
            return nums;  
        }  
    }  
}

哈哈，标准答案好精简阿，学习了。

日期：2018/2/8-12:31