【LeetCode】566. Reshape the Matrix

最新推荐文章于 2022-10-31 17:32:30 发布

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最新推荐文章于 2022-10-31 17:32:30 发布

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本文介绍了一个将原始矩阵通过行优先的方式重塑为指定尺寸的新矩阵的算法。若重塑不可能，则返回原矩阵。文章提供了具体实现代码及两个示例，帮助读者理解重塑过程。

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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:

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

这个题也不难。

不用考虑任何异常输入。

输入一个x*y 的矩阵，变成一个r*c的矩阵。

总体思路是x*y的矩阵先变成一个一维矩阵。再变成r*c的矩阵

var matrixReshape = function(nums, r, c) {
    var x=nums.length;
    var y=nums[0].length;
    var arr=[];
    var ret=[];
    if(x*y==r*c){
        for(var i=0;i<x;i++){
            arr=arr.concat(nums[i]);
        }
        for(i=0;i<r;i++){
          ret.push([]);
            for(var j=0;j<c;j++){
                ret[i][j]=arr[i*c+j];
            }
        }
        return ret;
    }else{
        return nums;
    }
};

效率也还不错~