车的可用捕获量（3.26leetcode每日打卡）

在一个 8 x 8 的棋盘上，有一个白色车（rook）。也可能有空方块，白色的象（bishop）和黑色的卒（pawn）。它们分别以字符 “R”，“.”，“B” 和 “p” 给出。大写字符表示白棋，小写字符表示黑棋。

车按国际象棋中的规则移动：它选择四个基本方向中的一个（北，东，西和南），然后朝那个方向移动，直到它选择停止、到达棋盘的边缘或移动到同一方格来捕获该方格上颜色相反的卒。另外，车不能与其他友方（白色）象进入同一个方格。

返回车能够在一次移动中捕获到的卒的数量。

 

输入：[[".",".",".",".",".",".",".","."],[".",".",".","p",".",".",".","."],[".",".",".","R",".",".",".","p"],[".",".",".",".",".",".",".","."],[".",".",".",".",".",".",".","."],[".",".",".","p",".",".",".","."],[".",".",".",".",".",".",".","."],[".",".",".",".",".",".",".","."]]
输出：3

输入：[[".",".",".",".",".",".",".","."],[".","p","p","p","p","p",".","."],[".","p","p","B","p","p",".","."],[".","p","B","R","B","p",".","."],[".","p","p","B","p","p",".","."],[".","p","p","p","p","p",".","."],[".",".",".",".",".",".",".","."],[".",".",".",".",".",".",".","."]]
输出：0
解释：
象阻止了车捕获任何卒。

 

思路：题目很简单就是简单的遍历四个方向即可，主要是好好审题，比如抓到一个卒就可停止，好好审题方能事半功倍。

 1 int numRookCaptures(char** board, int boardSize, int* boardColSize)
 2 {
 3     int cnt = 0;
 4     int row,col;
 5     int i,j;
 6
 7     //搜寻白色车所在位置
 8     //车 R
 9     //象 B
10     //卒 p
11     //空方格 .
12     for(i = 0; i < 8; i++)
13     {
14         for(j = 0; j < 8; j++)
15         {
16             if(board[i][j] == 'R')
17             {
18                 row = i;
19                 col = j;
20                 break;
21             }
22         }
23     }
24
25     //搜寻右边
26     for(i = col+1; i < 8; i++)
27     {
28         if(board[row][i] == 'p')
29         {
30             cnt++;
31             break;
32         }
33         if(board[row][i] == 'B')
34         break;
35     }
36
37     //搜寻左边
38     for(i = col-1; i >= 0; i--)
39     {
40         if(board[row][i] == 'p')
41         {
42             cnt++;
43             break;
44         }
45         if(board[row][i] == 'B')
46         break;
47     }
48
49     //搜寻下边
50     for(i = row+1; i < 8; i++)
51     {
52         if(board[i][col] == 'p')
53         {
54             cnt++;
55             break;
56         }
57         if(board[i][col] == 'B')
58         break;
59     }
60
61     //搜寻上边
62     for(i = row-1; i >= 0; i--)
63     {
64         if(board[i][col] == 'p')
65         {
66             cnt++;
67             break;
68         }
69         if(board[i][col] == 'B')
70         break;
71     }
72     return cnt;
73 }