[LeetCode每日一题]999. 可以被一步捕获的棋子数

ygxdsh

找到车然后朝四个方向延申，看看第一个棋是不是友军就行了

class Solution {
public:
    int numRookCaptures(vector<vector<char>>& board) {
        int ans=0;
        for(int i=0;i<8;i++)
        for(int j=0;j<8;j++)
        if(board[i][j]=='R')
        {
            int x1,x2,y1,y2;
            x1=x2=i,y1=y2=j;
            while(x1>0) 
            if(board[--x1][j]!='.')
            {
                if(board[x1][j]=='p') ans++;
                break;
            }
            while(x2<7) 
            if(board[++x2][j]!='.')
            {
                if(board[x2][j]=='p') ans++;
                break;
            }
            while(y1>0) 
            if(board[i][--y1]!='.')
            {
                if(board[i][y1]=='p') ans++;
                break;
            }
            while(y2<7) 
            if(board[i][++y2]!='.')
            {
                if(board[i][y2]=='p') ans++;
                break;
            }
            break;
        }
        return ans;
    }
};