LeetCode 999. 车的可用捕获量
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2022-05-23 18:39:54
我的LeetCode刷题源码[GitHub]:https://github.com/izhoujie/Algorithmcii LeetCode 999. 车的可用捕获量 题目 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 ......
我的leetcode刷题源码[github]:https://github.com/izhoujie/algorithmcii
leetcode 999. 车的可用捕获量
题目
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'
来源:力扣(leetcode)
链接:
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解题思路
思路1-题目较长,读懂后就简单了;
- 遍历找到r的位置;
- 从r的位置向四个方向上搜寻p,且p必须是首个遇到的字符,不能是b;
总结:本题的唯一难点就是题目又臭又长,读懂题比写代码要略难一点-。-
算法源码示例
package leetcode; /** * @author zhoujie * @date 2020年3月26日 下午1:21:57 * @description: 999. 车的可用捕获量 * */ public class leetcode_0999 { } class solution_0999 { /** * @author: zhoujie * @date: 2020年3月26日 下午1:38:42 * @param: @param board * @param: @return * @return: int * @description: 1- * */ public int numrookcaptures(char[][] board) { // 四个方向的增量 int[] x = new int[] { 1, -1, 0, 0 }; int[] y = new int[] { 0, 0, -1, 1 }; // 可捕获的目标数; int count = 0; for (int i = 0; i < 8; i++) { for (int j = 0; j < 8; j++) { // 找到r的位置 if (board[i][j] == 'r') { // 在四个方向上探寻p for (int k = 0; k < 4; k++) { int x1 = i + x[k]; int y1 = j + y[k]; while (x1 > -1 && x1 < 8 && y1 > -1 && y1 < 8) { if (board[x1][y1] == 'b') { break; } else if (board[x1][y1] == 'p') { count++; break; } else { x1 += x[k]; y1 += y[k]; } } } return count; } } } return count; } }