1. 题意

对角线排序

2. 题解

2.1 直接模拟

跟螺旋输出数组实际上有点像，这里需要枚举对角线数组的起始值。

class Solution {
public:
    void sort_dia(int br, int bc, vector<vector<int>> &mat) {
        vector<int> tmp;

        int r = mat.size();
        int c = mat[0].size();
        
        int cr = br;
        int cc = bc; 
        while ( cr < r && cc < c) {
            tmp.emplace_back( mat[cr][cc] );
            cr++,cc++;
        }
        sort(tmp.begin(), tmp.end());
        cr = br, cc = bc;

        int idx = 0;
        while ( cr < r && cc < c ) {
            mat[cr][cc] = tmp[idx];
            cr++,cc++,idx++;
        }
    }

    vector<vector<int>> diagonalSort(vector<vector<int>>& mat) {


        int r = mat.size();
        int c = mat[0].size();

        vector<vector<int>> ans;

        int cr = r - 1;
        int cc = 0;
        while (  cc < c) {
            if (cr) {
                sort_dia(cr, cc, mat);
                cr--;
                continue;
            }

            sort_dia(cr, cc, mat);
            cc++;
        }

        return vector<vector<int>>(mat);
    }
};

2.2 模拟+确定其实位置关系

同一对角线的位置

$i-j=k$,$k$为一定值。

所以我们只需要枚举$k$即可。

class Solution {
public:
    
    vector<vector<int>> diagonalSort(vector<vector<int>>& mat) {


        int r = mat.size();
        int c = mat[0].size();


        int br;
        int bc;

        vector<int> tmp(max(r,c), 0);
        // k = br - bc 
        // 
        for ( int k = 1 - c; k < r; ++k) {
            k >= 0 ? br = k : br = 0;
            bc = br - k ;
           
            int len = min(r - br, c - bc);
            
            for (int i = br; i < br + len;++i) {
                tmp[i - br] = mat[i][i - k];
            }
            sort(tmp.begin(), tmp.begin() + len);
            
            for (int i = br;i < br + len; ++i) {
                mat[i][ i - k ] = tmp[i - br];
            }
        }

        return vector<vector<int>>(mat);
    }
};

3. 参考

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