> For the complete documentation index, see [llms.txt](https://dfine.gitbook.io/leetcode/llms.txt). Markdown versions of documentation pages are available by appending `.md` to page URLs; this page is available as [Markdown](https://dfine.gitbook.io/leetcode/566.reshape_the_matrix.md).

# 566.Reshape the Matrix

**566.Reshape the Matrix**

难度:Easy

> 在MATLAB中，有一个非常有用的函数 reshape，它可以将一个矩阵重塑为另一个大小不同的新矩阵，但保留其原始数据。

给出一个由二维数组表示的矩阵，以及两个正整数r和c，分别表示想要的重构的矩阵的行数和列数。

重构后的矩阵需要将原始矩阵的所有元素以相同的行遍历顺序填充。

如果具有给定参数的reshape操作是可行且合理的，则输出新的重塑矩阵；否则，输出原始矩阵。

```
示例 1:

输入: 
nums = 
[[1,2],
 [3,4]]
r = 1, c = 4
输出: 
[[1,2,3,4]]
解释:
行遍历nums的结果是 [1,2,3,4]。新的矩阵是 1 * 4 矩阵, 用之前的元素值一行一行填充新矩阵。
示例 2:

输入: 
nums = 
[[1,2],
 [3,4]]
r = 2, c = 4
输出: 
[[1,2],
 [3,4]]
解释:
没有办法将 2 * 2 矩阵转化为 2 * 4 矩阵。 所以输出原矩阵。
注意：

给定矩阵的宽和高范围在 [1, 100]。
给定的 r 和 c 都是正数。
```

简单的遍历操作：

```
class Solution {
public:
    vector<vector<int>> matrixReshape(vector<vector<int>>& nums, int r, int c) {
        int r0=nums.size();
        int c0=nums[0].size();
        if(r0 * c0 != r*c) return nums;
        int sc=0,sr=0;
        vector<vector<int>> res(r,vector<int>(c,0));
        for(int i=0;i<r0;i++)
            for(int j=0;j<c0;j++)
            {
                
                res[sr][sc++]=nums[i][j];
                if(sc==c) {sr++; sc=0;} 
                
            }
        return res;
        
    }
};
```


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