题目如下:

On an 8x8 chessboard, there can be multiple Black Queens and one White King.

Given an array of integer coordinates queens that represents the positions of the Black Queens, and a pair of coordinates king that represent the position of the White King, return the coordinates of all the queens (in any order) that can attack the King.

Example 1:

Input: queens = [[0,1],[1,0],[4,0],[0,4],[3,3],[2,4]], king = [0,0]
Output: [[0,1],[1,0],[3,3]]
Explanation: 
The queen at [0,1] can attack the king cause they're in the same row.
The queen at [1,0] can attack the king cause they're in the same column.
The queen at [3,3] can attack the king cause they're in the same diagnal.
The queen at [0,4] can't attack the king cause it's blocked by the queen at [0,1].
The queen at [4,0] can't attack the king cause it's blocked by the queen at [1,0].
The queen at [2,4] can't attack the king cause it's not in the same row/column/diagnal as the king.

Example 2:

Input: queens = [[0,0],[1,1],[2,2],[3,4],[3,5],[4,4],[4,5]], king = [3,3]
Output: [[2,2],[3,4],[4,4]]

Example 3:

Input: queens = [[5,6],[7,7],[2,1],[0,7],[1,6],[5,1],[3,7],[0,3],[4,0],[1,2],[6,3],[5,0],[0,4],[2,2],[1,1],[6,4],
[5,4],[0,0],[2,6],[4,5],[5,2],[1,4],[7,5],[2,3],[0,5],[4,2],[1,0],[2,7],[0,1],[4,6],[6,1],[0,6],[4,3],[1,7]], king = [3,4]
Output: [[2,3],[1,4],[1,6],[3,7],[4,3],[5,4],[4,5]]

Constraints:

  • 1 <= queens.length <= 63
  • queens[0].length == 2
  • 0 <= queens[i][j] < 8
  • king.length == 2
  • 0 <= king[0], king[1] < 8
  • At most one piece is allowed in a cell.

解题思路:往king的上下左右,上左,上右,下左,下右八个方向移动,如果某个方向遇到queen,则表示这个queen能攻击到king,然后停止这个方向的移动。

代码如下:

class Solution(object):
def queensAttacktheKing(self, queens, king):
"""
:type queens: List[List[int]]
:type king: List[int]
:rtype: List[List[int]]
"""
res = []
dic = {}
for (x,y) in queens:dic[(x,y)] = 1
direction = [(0,1),(0,-1),(-1,0),(1,0),(-1,-1),(-1,1),(1,-1),(1,1)]
for (x,y) in direction:
kx,ky = king
while kx + x >= 0 and kx + x < 8 and ky+y>=0 and ky+ y < 8:
if (kx+x,ky+y) in dic:
res.append([kx+x,ky+y])
break
kx += x
ky += y
return res

【leetcode】1222. Queens That Can Attack the King的更多相关文章

  1. 【LeetCode】1222. Queens That Can Attack the King 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 遍历 日期 题目地址:https://leetcode ...

  2. 【LeetCode】Minimum Depth of Binary Tree 二叉树的最小深度 java

    [LeetCode]Minimum Depth of Binary Tree Given a binary tree, find its minimum depth. The minimum dept ...

  3. 【Leetcode】Pascal&#39;s Triangle II

    Given an index k, return the kth row of the Pascal's triangle. For example, given k = 3, Return [1,3 ...

  4. 53. Maximum Subarray【leetcode】

    53. Maximum Subarray[leetcode] Find the contiguous subarray within an array (containing at least one ...

  5. 27. Remove Element【leetcode】

    27. Remove Element[leetcode] Given an array and a value, remove all instances of that value in place ...

  6. 【刷题】【LeetCode】007-整数反转-easy

    [刷题][LeetCode]总 用动画的形式呈现解LeetCode题目的思路 参考链接-空 007-整数反转 方法: 弹出和推入数字 & 溢出前进行检查 思路: 我们可以一次构建反转整数的一位 ...

  7. 【刷题】【LeetCode】000-十大经典排序算法

    [刷题][LeetCode]总 用动画的形式呈现解LeetCode题目的思路 参考链接 000-十大经典排序算法

  8. 【leetcode】893. Groups of Special-Equivalent Strings

    Algorithm [leetcode]893. Groups of Special-Equivalent Strings https://leetcode.com/problems/groups-o ...

  9. 【leetcode】657. Robot Return to Origin

    Algorithm [leetcode]657. Robot Return to Origin https://leetcode.com/problems/robot-return-to-origin ...

随机推荐

  1. java游戏服务器--简单工厂模式

    先来学习下简单工厂模式! 我们知道在游戏里有很多的场景,例如:帮派场景,副本场景,野外场景... 现在我们有这样的需求: 1.我们需要进入帮派场景时---开始执行帮派任务. 2.我们需要进入副本场景时 ...

  2. qrcode-reader——二维码识别

    JavaScript QRCode reader for HTML5 enabled browser 参考资料1:[https://www.npmjs.com/package/qrcode-reade ...

  3. 在Spring容器外调用bean

    这个东西源于这种需求:一个应用丢到服务其后,不管用户有没有访问项目,这个后台线程都必须给我跑,而且这个线程还调用了Spring注入的bean,这样自然就会想到去监听Servlet的状态,当Servle ...

  4. beego框架学习(二) -路由设置

    路由设置 什么是路由设置呢?前面介绍的 MVC 结构执行时,介绍过 beego 存在三种方式的路由:固定路由.正则路由.自动路由,接下来详细的讲解如何使用这三种路由. 基础路由 从beego1.2版本 ...

  5. PHP 按照时区获取当前时间

    /**  * 时间格式化  * @param string $dateformat 时间格式  * @param int $timestamp 时间戳  * @param int $timeoffse ...

  6. 微信小程序购物车实现

    1,wxml <view class="miniCart-wrap {{isIpx?'is-ipx':''}}"> <view class="miniC ...

  7. python 并发编程 多线程 GIL与多线程

    GIL与多线程 有了GIL的存在,同一时刻同一进程中只有一个线程被执行 多进程可以利用多核,但是开销大,而python的多线程开销小,但却无法利用多核优势 1.cpu到底是用来做计算的,还是用来做I/ ...

  8. CDH6.2中capacity队列的分配

    配置: yarn.scheduler.capacity.root.queues

  9. 来自 Vue 3.0 的 Composition API 尝鲜

    来自 Vue 3.0 的 Composition API 尝鲜:https://segmentfault.com/a/1190000020205747

  10. 洛谷 P2384 最短路 题解

    题面 这道题需要用到一个神奇的知识点:log(n*m)=log(n)+log(m): 所以对所有边权取个log,然后算log的最短路的同时维护乘积即可 #include <bits/stdc++ ...