题解 | #棋子翻转#

#include <utility>
#include <vector>
class Solution {
  public:
    /**
     * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
     *
     *
     * @param A int整型vector<vector<>>
     * @param f int整型vector<vector<>>
     * @return int整型vector<vector<>>
     */
    vector<vector<int> > flipChess(vector<vector<int> >& A,
                                   vector<vector<int> >& f) {

        for (int i = 0 ; i < f.size() ; i++ ) {
            pair <int, int> p ;
            p = make_pair(f[i][0], f[i][1]) ;
            reverse(A, p) ;
        }

        return A ;
    }

    void reverse(vector<vector<int>>& A, pair<int, int> p ) {
        int dx[4] = {1, 0, -1, 0} ;
        int dy[4] = {0, 1, 0, -1} ;
        // A是棋盘
        for (int i = 0 ; i < A.size()  ; i++) {
            for (int j = 0 ; j < A[i].size() ; j++ ) {
                if (i == (p.first - 1) && (j == p.second - 1 ) ) {
                    for (int k = 0 ; k < 4 ; k++) {
                        int d_x = i + dx[k] ;
                        int d_y = j + dy[k] ;
                        //cout << d_x << " " << d_y << endl ;
                        if (d_x < A.size() && d_y < A[i].size()) {

                            if (A[d_x][d_y] == 1 ) {
                                A[d_x][d_y] = 0 ;
                            } else {
                                A[d_x][d_y] = 1 ;
                            }
                        }


                    }
                }
            }
        }

    }
};