999. 车的可用捕获量
难度简单
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。
示例 1:
输入:[[".",".",".",".",".",".",".","."],[".",".",".","p",".",".",".","."],[".",".",".","R",".",".",".","p"],[".",".",".",".",".",".",".","."],[".",".",".",".",".",".",".","."],[".",".",".","p",".",".",".","."],[".",".",".",".",".",".",".","."],[".",".",".",".",".",".",".","."]] 输出:3 解释: 在本例中,车能够捕获所有的卒。
示例 2:
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。
示例 3:
输入:[[".",".",".",".",".",".",".","."],[".",".",".","p",".",".",".","."],[".",".",".","p",".",".",".","."],["p","p",".","R",".","p","B","."],[".",".",".",".",".",".",".","."],[".",".",".","B",".",".",".","."],[".",".",".","p",".",".",".","."],[".",".",".",".",".",".",".","."]] 输出:3 解释: 车可以捕获位置 b5,d6 和 f5 的卒。
提示:
board.length == board[i].length == 8
board[i][j]
可以是'R'
,'.'
,'B'
或'p'
- 只有一个格子上存在
board[i][j] == 'R'
思路:刚开始首先想到的是深度搜索,后来发现用四个for就可以解决这个问题,首先遍历数组,找到车的位置,然后向四个方向进行搜索,就可以得出答案。
int numRookCaptures(char** board, int boardSize, int* boardColSize){ int X, Y; int cnt = 0, i, j; for (i = 0; i < 8; i++) { for (j = 0; j < 8; j++) { if (board[i][j] == 'R') { X = i; Y = j; } } } for (i = Y; i >= 0; i--) { if (board[X][i] == 'p') { cnt++; break; } if (board[X][i] == 'B') { break; } } for (i = Y; i < 8; i++) { if (board[X][i] == 'p') { cnt++; break; } if (board[X][i] == 'B') { break; } } for (i = X; i >= 0; i--) { if (board[i][Y] == 'p') { cnt++; break; } if (board[i][Y] == 'B') { break; } } for (i = X; i < 8; i++) { if (board[i][Y] == 'p') { cnt++; break; } if (board[i][Y] == 'B') { break; } } return cnt; }