> 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/1329.sort_the_matrix_diagonally.md).

# 1329.Sort the Matrix Diagonally

**1329.Sort the Matrix Diagonally**

难度:Medium

> 给你一个 m \* n 的整数矩阵 mat ，请你将同一条对角线上的元素（从左上到右下）按升序排序后，返回排好序的矩阵。

Example:\
![](https://i.loli.net/2020/04/30/7dhQPSZztT3WMuH.png)

```
输入：mat = [[3,3,1,1],[2,2,1,2],[1,1,1,2]]
输出：[[1,1,1,1],[1,2,2,2],[1,2,3,3]]
```

提示：

```
m == mat.length
n == mat[i].length
1 <= m, n <= 100
1 <= mat[i][j] <= 100
```

思路： 分两种情况，一种是宽大于高，一种是宽小于高。 排序分三步，第一部分是左下部分的对角线排序，第二部分是右上部分的对角线排序，第三部分是中间部分对角线排序。

```
class Solution {
public:
    vector<vector<int>> diagonalSort(vector<vector<int>>& mat) {
        int m=mat.size();
        int n=mat[0].size();
 
        
        if(m>n)
        {
        for(int k=m-2;k>m-n;k--)
        {
            for(int i=k;i<m-1;i++)
            for(int j=i+1;j<m;j++)
            {
                if(mat[i][i-k]>mat[j][j-k])
                swap(mat[i][i-k], mat[j][j-k]);
            }   
        } 
        for(int k=n-2;k>0;k--)
        {    
            for(int i=k;i<n-1;i++)
            for(int j=i+1;j<n;j++)
            {
                if(mat[i-k][i]>mat[j-k][j])
                swap(mat[i-k][i], mat[j-k][j]);
            }        

        }

             for(int k=0;k<=m-n;k++)
            {
                for(int i=0;i<n-1;i++)
                for(int j=i+1;j<n;j++)
                if(mat[i+k][i]>mat[j+k][j])
                swap(mat[i+k][i], mat[j+k][j]);
            } 
        }
        else 
        {
        for(int k=m-2;k>0;k--)
        {
            for(int i=k;i<m-1;i++)
            for(int j=i+1;j<m;j++)
            {
                if(mat[i][i-k]>mat[j][j-k])
                swap(mat[i][i-k], mat[j][j-k]);
            }   
        } 
        for(int k=n-m+1;k<n-1;k++)
        {    
            for(int i=k;i<n-1;i++)
            for(int j=i+1;j<n;j++)
            {
                if(mat[i-k][i]>mat[j-k][j])
                swap(mat[i-k][i], mat[j-k][j]);
            }        

        }
            for(int k=0;k<=n-m;k++)
            {
                for(int i=0;i<m-1;i++)
                for(int j=i+1;j<m;j++)
                if(mat[i][i+k]>mat[j][j+k])
                swap(mat[i][i+k], mat[j][j+k]);
            } 
        }
        return mat;

    }
};
```

> 执行用时 :24 ms, 在所有 C++ 提交中击败了21.23%的用户\
> 内存消耗 :8.4 MB, 在所有 C++ 提交中击败了100.00%的用户


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