【LeetCode】Available Captures for Rook(车的可用捕获量)
2024-09-30 06:49:05
这道题是LeetCode里的第999道题。
题目叙述:
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'
这道题很简单,首先我们要找到车的位置在哪,然后从车这个位置开始遍历它的上下左右方向的格子,判断是否为象或者卒。
代码如下:
class Solution {
public int numRookCaptures(char[][] board) {
int R_x=-1,R_y=-1;
int res=0;
//find R
for(int i=0;i<8;i++) {
for(int j=0;j<8;j++) {
if(board[i][j]=='R') {
R_x=i;R_y=j;break;
}
}
if(R_x>=0 && R_y>=0)break;
}
//up,down,left,right
for(int i=1;i+R_y<8;i++) {
if(board[R_x][i+R_y]=='B')break;
else if(board[R_x][i+R_y]=='p'){res++;break;}
}
for(int i=1;R_y-i>=0;i++) {
if(board[R_x][R_y-i]=='B')break;
else if(board[R_x][R_y-i]=='p') {res++;break;}
}
for(int i=1;R_x+i<8;i++) {
if(board[R_x+i][R_y]=='B')break;
else if(board[R_x+i][R_y]=='p') {res++;break;}
}
for(int i=1;R_x-i>=0;i++) {
if(board[R_x-i][R_y]=='B')break;
else if(board[R_x-i][R_y]=='p'){res++;break;}
}
return res;
}
}
提交结果:
个人总结:
这题数组题真的太简单了,被第1000题虐后来这里找找自信!
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