[LeetCode] 723. Candy Crush 糖果粉碎
This question is about implementing a basic elimination algorithm for Candy Crush.
Given a 2D integer array board
representing the grid of candy, different positive integers board[i][j]
represent different types of candies. A value of board[i][j] = 0
represents that the cell at position (i, j)
is empty. The given board represents the state of the game following the player's move. Now, you need to restore the board to a stable state by crushing candies according to the following rules:
- If three or more candies of the same type are adjacent vertically or horizontally, "crush" them all at the same time - these positions become empty.
- After crushing all candies simultaneously, if an empty space on the board has candies on top of itself, then these candies will drop until they hit a candy or bottom at the same time. (No new candies will drop outside the top boundary.)
- After the above steps, there may exist more candies that can be crushed. If so, you need to repeat the above steps.
- If there does not exist more candies that can be crushed (ie. the board is stable), then return the current board.
You need to perform the above rules until the board becomes stable, then return the current board.
Example 1:
Input:
board =
[[110,5,112,113,114],[210,211,5,213,214],[310,311,3,313,314],[410,411,412,5,414],[5,1,512,3,3],[610,4,1,613,614],[710,1,2,713,714],[810,1,2,1,1],[1,1,2,2,2],[4,1,4,4,1014]]
Output:
[[0,0,0,0,0],[0,0,0,0,0],[0,0,0,0,0],[110,0,0,0,114],[210,0,0,0,214],[310,0,0,113,314],[410,0,0,213,414],[610,211,112,313,614],[710,311,412,613,714],[810,411,512,713,1014]]
Explanation:
Note:
- The length of
board
will be in the range [3, 50]. - The length of
board[i]
will be in the range [3, 50]. - Each
board[i][j]
will initially start as an integer in the range [1, 2000].
一个类似糖果消消乐游戏的题,给一个二维数组board,数组代表糖果,在水平和竖直方向上3个或以上连续相同的数字可以被消除,消除后位于上方的数字会填充空位。注意是所以能消除的糖果消除后然后再落下,进行第二次消除。直到最后没有能消除的了,达到稳定状态。
解法:标记出所有需要被crush的元素,然后去掉这些元素,将上面的数下降,空位变为0。难的地方是如何确定哪些元素需要被crush。
Python:
class Solution(object):
def candyCrush(self, board):
"""
:type board: List[List[int]]
:rtype: List[List[int]]
"""
R, C = len(board), len(board[0])
changed = True while changed:
changed = False for r in xrange(R):
for c in xrange(C-2):
if abs(board[r][c]) == abs(board[r][c+1]) == abs(board[r][c+2]) != 0:
board[r][c] = board[r][c+1] = board[r][c+2] = -abs(board[r][c])
changed = True for r in xrange(R-2):
for c in xrange(C):
if abs(board[r][c]) == abs(board[r+1][c]) == abs(board[r+2][c]) != 0:
board[r][c] = board[r+1][c] = board[r+2][c] = -abs(board[r][c])
changed = True for c in xrange(C):
i = R-1
for r in reversed(xrange(R)):
if board[r][c] > 0:
board[i][c] = board[r][c]
i -= 1
for r in reversed(xrange(i+1)):
board[r][c] = 0 return board
C++:
// Time: O((R * C)^2)
// Space: O(1) class Solution {
public:
vector<vector<int>> candyCrush(vector<vector<int>>& board) {
const auto R = board.size(), C = board[0].size();
bool changed = true; while (changed) {
changed = false; for (int r = 0; r < R; ++r) {
for (int c = 0; c + 2 < C; ++c) {
auto v = abs(board[r][c]);
if (v != 0 && v == abs(board[r][c + 1]) && v == abs(board[r][c + 2])) {
board[r][c] = board[r][c + 1] = board[r][c + 2] = -v;
changed = true;
}
}
} for (int r = 0; r + 2 < R; ++r) {
for (int c = 0; c < C; ++c) {
auto v = abs(board[r][c]);
if (v != 0 && v == abs(board[r + 1][c]) && v == abs(board[r + 2][c])) {
board[r][c] = board[r + 1][c] = board[r + 2][c] = -v;
changed = true;
}
}
} for (int c = 0; c < C; ++c) {
int empty_r = R - 1;
for (int r = R - 1; r >= 0; --r) {
if (board[r][c] > 0) {
board[empty_r--][c] = board[r][c];
}
}
for (int r = empty_r; r >= 0; --r) {
board[r][c] = 0;
}
}
} return board;
}
};
C++:
class Solution {
public:
vector<vector<int>> candyCrush(vector<vector<int>>& board) {
int m = board.size(), n = board[0].size();
while (true) {
vector<pair<int, int>> del;
for (int i = 0; i < m; ++i) {
for (int j = 0; j < n; ++j) {
if (board[i][j] == 0) continue;
int x0 = i, x1 = i, y0 = j, y1 = j;
while (x0 >= 0 && x0 > i - 3 && board[x0][j] == board[i][j]) --x0;
while (x1 < m && x1 < i + 3 && board[x1][j] == board[i][j]) ++x1;
while (y0 >= 0 && y0 > j - 3 && board[i][y0] == board[i][j]) --y0;
while (y1 < n && y1 < j + 3 && board[i][y1] == board[i][j]) ++y1;
if (x1 - x0 > 3 || y1 - y0 > 3) del.push_back({i, j});
}
}
if (del.empty()) break;
for (auto a : del) board[a.first][a.second] = 0;
for (int j = 0; j < n; ++j) {
int t = m - 1;
for (int i = m - 1; i >= 0; --i) {
if (board[i][j]) swap(board[t--][j], board[i][j]);
}
}
}
return board;
}
};
C++:
class Solution {
public:
vector<vector<int>> candyCrush(vector<vector<int>>& board) {
int m = board.size(), n = board[0].size();
bool toBeContinued = false; for (int i = 0; i < m; ++i) { // horizontal crushing
for (int j = 0; j + 2 < n; ++j) {
int& v1 = board[i][j];
int& v2 = board[i][j+1];
int& v3 = board[i][j+2]; int v0 = std::abs(v1); if (v0 && v0 == std::abs(v2) && v0 == std::abs(v3)) {
v1 = v2 = v3 = - v0;
toBeContinued = true;
}
}
} for (int i = 0; i + 2 < m; ++i) { // vertical crushing
for (int j = 0; j < n; ++j) {
int& v1 = board[i][j];
int& v2 = board[i+1][j];
int& v3 = board[i+2][j];
int v0 = std::abs(v1);
if (v0 && v0 == std::abs(v2) && v0 == std::abs(v3)) {
v1 = v2 = v3 = -v0;
toBeContinued = true;
}
}
} for (int j = 0; j < n; ++j) { // gravity step
int dropTo = m - 1;
for (int i = m - 1; i >= 0; --i) {
if (board[i][j] >= 0) {
board[dropTo--][j] = board[i][j];
}
} for (int i = dropTo; i >= 0; i--) {
board[i][j] = 0;
}
} return toBeContinued ? candyCrush(board) : board; }
};
All LeetCode Questions List 题目汇总
[LeetCode] 723. Candy Crush 糖果粉碎的更多相关文章
- [LeetCode] 723. Candy Crush 糖果消消乐
This question is about implementing a basic elimination algorithm for Candy Crush. Given a 2D intege ...
- LeetCode 723. Candy Crush
原题链接在这里:https://leetcode.com/problems/candy-crush/ 题目: This question is about implementing a basic e ...
- [LeetCode] Candy Crush 糖果消消乐
This question is about implementing a basic elimination algorithm for Candy Crush. Given a 2D intege ...
- 【LeetCode】723. Candy Crush 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 暴力 日期 题目地址:https://leetcode ...
- [LeetCode] Candy (分糖果),时间复杂度O(n),空间复杂度为O(1),且只需遍历一次的实现
[LeetCode] Candy (分糖果),时间复杂度O(n),空间复杂度为O(1),且只需遍历一次的实现 原题: There are N children standing in a line. ...
- LeetCode:135. 分发糖果
LeetCode:135. 分发糖果 老师想给孩子们分发糖果,有 N 个孩子站成了一条直线,老师会根据每个孩子的表现,预先给他们评分. 你需要按照以下要求,帮助老师给这些孩子分发糖果: 每个孩子至少分 ...
- [LeetCode] Candy 分糖果问题
There are N children standing in a line. Each child is assigned a rating value. You are giving candi ...
- LeetCode OJ:Candy(糖果问题)
There are N children standing in a line. Each child is assigned a rating value. You are giving candi ...
- LeetCode.888-公平的糖果交换(Fair Candy Swap)
这是悦乐书的第339次更新,第363篇原创 01 看题和准备 今天介绍的是LeetCode算法题中Easy级别的第208题(顺位题号是888).Alice和Bob有不同大小的糖果棒:A[i]是Alic ...
随机推荐
- gson之将对象转化成json字符串的方法
public class GsonUtil { /** * 将object对象转成json格式字符串 */ public static String toJson(Object object) { G ...
- 注解@Transient
@Transient表示该属性并非一个到数据库表的字段的映射,ORM框架将忽略该属性. 如果一个属性并非数据库表的字段映射,就务必将其标示为@Transient,否则,ORM框架默认其注解为@Bas ...
- 项目Alpha冲刺 8
作业描述 课程: 软件工程1916|W(福州大学) 作业要求: 项目Alpha冲刺(团队) 团队名称: 火鸡堂 作业目标: 介绍第8天冲刺的项目进展.问题困难和心得体会 1.团队信息 队名:火鸡堂 队 ...
- Centos7 Mysql主从双机热备的实战记录
操作系统:centos7 服务器:主:172.19.242.69 从:172.19.242.70 1. 服务器 由于Mysql不同版本之间的(二进制日志)binlog格式可能会不太一样,因此最好的 ...
- 【python】Requests的三种参数请求方式
URL参数请求: import requests ''' URL Parameters 请求方式: URL参数 例如: 以get 方式请求http://httpbin.org/get?first_na ...
- 【python】json中load和loads区别
相同点 dump 和 dumps 都实现了序列化 load 和 loads 都实现反序列化 变量从内存中变成可存储或传输的过程称之为序列化序列化是将对象状态转化为可保存或可传输格式的过程. 变量内容从 ...
- HttpContext对象下的属性Application、Cache、Request、Response、Server、Session、User
概述: HttpContext封装关于单个HTTP请求的所有HTTP特定信息. HttpContext基于HttpApplication的处理管道,由于HttpContext对象贯穿整个处理过程,所以 ...
- PHP-FPM config 文件生产环境
;;;;;;;;;;;;;;;;;; ; Global Options ; ;;;;;;;;;;;;;;;;;; [global] pid = run/php-fpm.pid error_log = ...
- 产品生命周期(Product Life Circle,PLC)
什么是产品生命周期? 产品生命周期是新产品从开发进入市场到被市场淘汰的整个过程.产品生命周期可分为初创期.成长期.成熟期.衰退期. 产品生命周期有什么用? 在产品不同的生命阶段,公司的业务目的都不同. ...
- Time of Trial
Time of Trial(思维) 题目大意:一个人想逃,一个人要阻止那个人逃跑 ,阻止者念每一个符咒的时间为b[i],逃跑者念每一个符咒的时间为a[i],逃跑者如果念完了第i个符咒,阻止者还没有念完 ...