leetcode- 距离顺序排序矩阵单元格

C++解法：

#include <iostream>
#include <vector>
#include <map>
#include <algorithm>

using namespace std;

class Solution {
public:
    vector<vector<int>> allCellsDistOrder(int R, int C, int r0, int c0) {
        vector<vector<int>> res(R*C, vector<int>());//数组第三个位置存放两点的曼哈顿距离
        int num = ;
        for (int i = ; i<C; i++)
        {
            for (int j = ; j<R; j++)
            {
                res[num][] = j;
                res[num][] = i;
                res[num][] = abs(r0 - j) + abs(c0 - i);//曼哈顿距离
                num++;
            }
        }
        sort(res.begin(), res.end(), ismax);//排序
        for (int i = ; i<num; i++)//将曼哈顿距离删除
        {
            res[i].pop_back();
        }
        return res;
    }
    static bool ismax(vector<int> &a, vector<int> &b)//根据曼哈顿距离升序排序
    {
        return a[]<b[];
    }
};

int main()
{
    Solution m;
    m.allCellsDistOrder(, , , );

    return ;
}

Python解法：

def allCellsDistOrder(R, C, r0, c0):
    dist_list = [[] for i in range(R+C)]
    for i in range(R):
        for j in range(C):
            distinct = abs(r0 - i) + abs(c0 - j)
            dist_list[distinct].append([i, j])
    result = []
    for i in dist_list:
        if i:
            result.extend(i)
        else:
            break
    return result

print(allCellsDistOrder(2, 3, 1, 2))