作者: 负雪明烛
id: fuxuemingzhu
个人博客: http://fuxuemingzhu.cn/


题目地址:https://leetcode.com/problems/available-captures-for-rook/

题目描述

On an 8 x 8 chessboard, there is one white rook. There also may be empty squares, white bishops, and black pawns. These are given as characters ‘R’, ‘.’, ‘B’, and ‘p’ respectively. Uppercase characters represent white pieces, and lowercase characters represent black pieces.

The rook moves as in the rules of Chess: it chooses one of four cardinal directions (north, east, west, and south), then moves in that direction until it chooses to stop, reaches the edge of the board, or captures an opposite colored pawn by moving to the same square it occupies. Also, rooks cannot move into the same square as other friendly bishops.

Return the number of pawns the rook can capture in one move.

Example 1:

Input: [[".",".",".",".",".",".",".","."],[".",".",".","p",".",".",".","."],[".",".",".","R",".",".",".","p"],[".",".",".",".",".",".",".","."],[".",".",".",".",".",".",".","."],[".",".",".","p",".",".",".","."],[".",".",".",".",".",".",".","."],[".",".",".",".",".",".",".","."]]
Output: 3
Explanation:
In this example the rook is able to capture all the pawns.

Example 2:

Input: [[".",".",".",".",".",".",".","."],[".","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",".","."],[".",".",".",".",".",".",".","."],[".",".",".",".",".",".",".","."]]
Output: 0
Explanation:
Bishops are blocking the rook to capture any pawn.

Example 3:

Input: [[".",".",".",".",".",".",".","."],[".",".",".","p",".",".",".","."],[".",".",".","p",".",".",".","."],["p","p",".","R",".","p","B","."],[".",".",".",".",".",".",".","."],[".",".",".","B",".",".",".","."],[".",".",".","p",".",".",".","."],[".",".",".",".",".",".",".","."]]
Output: 3
Explanation:
The rook can capture the pawns at positions b5, d6 and f5.

Note:

  1. board.length == board[i].length == 8
  2. board[i][j] is either ‘R’, ‘.’, ‘B’, or ‘p’
  3. There is exactly one cell with board[i][j] == ‘R’

题目大意

在一个国际象棋的棋盘上,有一个白车(R),有若干白象(B)、黑卒(p),其余是空白(.),问这个白车在只移动一次的情况下,能吃掉哪几个黑卒。

解题方法

暴力遍历

棋盘只有8*8,只有一个白车,所以做法可以很简单地从白车出发,向四个方向进行搜索即可!

C++代码如下:

class Solution {
public:
int numRookCaptures(vector<vector<char>>& board) {
const int N = 8;
pair<int, int> pos;
for (int i = 0; i < N; ++i) {
for (int j = 0; j < N; ++j) {
if (board[i][j] == 'R') {
pos.first = i;
pos.second = j;
break;
}
}
}
vector<vector<int>> dirs = {{0, 1}, {0, -1}, {1, 0}, {-1, 0}};
int res = 0;
for (int i = pos.first + 1; i < N; ++i) {
if (board[i][pos.second] == 'B')
break;
if (board[i][pos.second] == 'p') {
++res;
break;
}
}
for (int i = pos.first - 1; i >= 0; --i) {
if (board[i][pos.second] == 'B')
break;
if (board[i][pos.second] == 'p') {
++res;
break;
}
}
for (int j = pos.second + 1; j < N; ++j) {
if (board[pos.first][j] == 'B')
break;
if (board[pos.first][j] == 'p') {
++res;
break;
}
}
for (int j = pos.second - 1; j >= 0; --j) {
if (board[pos.first][j] == 'B')
break;
if (board[pos.first][j] == 'p') {
++res;
break;
}
}
return res;
}
};

日期

2019 年 2 月 24 日 —— 周末又结束了

【LeetCode】999. Available Captures for Rook 解题报告(C++)的更多相关文章

  1. 【LeetCode】999. Available Captures for Rook 解题报告(C++)

    作者: 负雪明烛 id: fuxuemingzhu 个人博客:http://fuxuemingzhu.cn/ 目录 题目描述 题目大意 解题方法 四方向搜索 日期 题目地址:https://leetc ...

  2. Leetcode 999. Available Captures for Rook

    class Solution: def numRookCaptures(self, board: List[List[str]]) -> int: rook = [0, 0] ans = 0 f ...

  3. 【LEETCODE】46、999. Available Captures for Rook

    package y2019.Algorithm.array; /** * @ProjectName: cutter-point * @Package: y2019.Algorithm.array * ...

  4. 【LeetCode】760. Find Anagram Mappings 解题报告

    [LeetCode]760. Find Anagram Mappings 解题报告 标签(空格分隔): LeetCode 题目地址:https://leetcode.com/problems/find ...

  5. 【LeetCode】Pascal's Triangle II 解题报告

    [LeetCode]Pascal's Triangle II 解题报告 标签(空格分隔): LeetCode 题目地址:https://leetcode.com/problems/pascals-tr ...

  6. 【LeetCode】299. Bulls and Cows 解题报告(Python)

    [LeetCode]299. Bulls and Cows 解题报告(Python) 作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu.cn/ 题 ...

  7. 【LeetCode】743. Network Delay Time 解题报告(Python)

    [LeetCode]743. Network Delay Time 解题报告(Python) 标签(空格分隔): LeetCode 作者: 负雪明烛 id: fuxuemingzhu 个人博客: ht ...

  8. 【LeetCode】518. Coin Change 2 解题报告(Python)

    [LeetCode]518. Coin Change 2 解题报告(Python) 作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu.cn/ 题目 ...

  9. 【LeetCode】474. Ones and Zeroes 解题报告(Python)

    [LeetCode]474. Ones and Zeroes 解题报告(Python) 作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu.cn/ ...

随机推荐

  1. perl 获取目录信息

    1 #!/usr/bin/perl -w 2 use strict; 3 use FindBin qw($Bin $Script); 4 5 my $rp=$Bin; 6 print "th ...

  2. Go语言缺陷

    我为什么放弃Go语言 目录(?)[+] 我为什么放弃Go语言 有好几次,当我想起来的时候,总是会问自己:我为什么要放弃Go语言?这个决定是正确的吗?是明智和理性的吗?其实我一直在认真思考这个问题. 开 ...

  3. Vector总结及部分底层源码分析

    Vector总结及部分底层源码分析 1. Vector继承的抽象类和实现的接口 Vector类实现的接口 List接口:里面定义了List集合的基本接口,Vector进行了实现 RandomAcces ...

  4. IPv6 私有地址

    在互联网的地址架构中,专用网络是指遵守RFC 1918(IPV4)和RFC 4193(IPV6)规范,使用专用IP地址空间的网络.私有IP无法直接连接互联网,需要使用网络地址转换(Network Ad ...

  5. 什么是 IP 地址 – 定义和解释

    IP 地址定义 IP 地址是一个唯一地址,用于标识互联网或本地网络上的设备.IP 代表"互联网协议",它是控制通过互联网或本地网络发送的数据格式的一组规则. 本质上,IP 地址是允 ...

  6. Oracle中常用的系统函数

    本文主要来梳理下Oracle中的常用的系统函数,掌握这些函数的使用,对于我们编写SQL语句或PL/SQL代码时很有帮助,所以这也是必须掌握的知识点. 本文主要包括以下函数介绍:1.字符串函数2. 数值 ...

  7. soapui pro 5.1.2 的破解方法

    Protection-4.6,和scz.key这两个文件能破解5.1.2的SoapUI 的Pro版本,mac 和 windows均可.1.拷贝Protection-4.6.jar到soapui安装的l ...

  8. 【编程思想】【设计模式】【行为模式Behavioral】registry

    Python版 https://github.com/faif/python-patterns/blob/master/behavioral/registry.py #!/usr/bin/env py ...

  9. String直接赋值与使用new String的区别

    在研究String直接赋值与new String的区别之前我们需要先了解java中的字符串常量池的概念 字符串常量池 String类是我们平常项目中使用频率非常高的一种对象类型,jvm为了提升性能和减 ...

  10. 3.使用Spring Data ElasticSearch操作ElasticSearch(5.6.8版本)

    1.引入maven坐标 <!--spring-data-elasticsearch--><dependency> <groupId>org.springframew ...