Lintcode 3656 · Design Snake Game (贪吃蛇,设计题)

3656 · Design Snake Game

Medium

Description

In this question, you need to design a snake game class SnakeGame.

Snake Game: There exists a snake in a matrix. Food will appear continuously in this matrix, and we need to control the snake to move continuously (through the four directions of up, down, left, and right) to eat the food. after eating food. The length of the snake will increase by one bit. At the same time, the game score will be increased by 1. Afterwards, the next food will be refreshed (the position occupied by the snake will not be refreshed). If the snake hits the wall or eats its own body while eating the food, the game is over.

The function's constructor receives the following parameters:

width: the number of columns in the snake matrix interface

height: the number of rows in the snake matrix interface

foods: The coordinates of the food, represented by a two-dimensional array, where the first element of each array is the row coordinate, and the second element is the column coordinate

In addition, you need to complete the move() function, which will receive a string indicating the direction. Indicates that the snake needs to move one space in this direction. You need to return the Snake score after the move (initial score is 0). Returns -1 if the game ends after the snake moves.

Only $39.9 for the "Twitter Comment System Project Practice" within a limited time of 7 days!

WeChat Notes Twitter for more information(WeChat ID jiuzhang104)

Example

Example 1

Input:

3

2

\[1, 2\], \[0, 0\]

"r", "d", "r", "u", "l", "l", "l"

Output:

0, 0, 1, 1, 1, 1, 2, -1

Explanation:

The initial state is as follows:

s | |

| | f

Among them, s represents the snake, and f represents the current food.

After action "r":

| s |

| | f

...

After eating the first food:

f | |

| s |s

...

After eating the second food:

s | s | s

| |

Example 2

Input:

3

2

\[0, 1\], \[1, 1\], \[1, 0\], \[0, 2\]

"r", "d", "l", "u", "r", "r", "d", "l", "u"

Output:

1, 2, 3, 3, 3, 4, 4, 4, -1

解法1:蛇的身体用deque。另外再用set判重(如果吃到自己的身体的话)。

食物可以用queue,因为是按先后顺序一个一个出现。

为什么蛇的身体用deque呢,因为吃到食物要pop_tail,而在游走的过程中要pop_front。

cpp 复制代码
class SnakeGame {
public:
    SnakeGame(int width, int height, vector<vector<int>>& foods) : width(width), height(height) {
        for (auto food : foods) {
            foodQueue.push({food[0], food[1]});
        }
        currPos = {0, 0}; //initial position (0,0)
        score = 0;
    }

    /**
     * @param direction: the direction of the move
     * @return: the score after the move
     */
    int move(string &direction) {
        int n = direction.size();
        switch (direction[0]) {
            case 'r':
                currPos.second++;
                break;
            case 'l':
                currPos.second--;
                break;
            case 'd':
                currPos.first++;
                break;
            case 'u':
                currPos.first--;
                break;
        }
        if (currPos.first < 0 || currPos.first >= height) return -1;
        if (currPos.second < 0 || currPos.second >= width) return -1;
        if (bodySet.find(currPos) != bodySet.end()) return -1;
        bodySet.insert(currPos);
        bodyQueue.push_back(currPos);
        if (!foodQueue.empty() && currPos == foodQueue.front()) {
            foodQueue.pop();
            score++;
        } else {
            bodySet.erase(bodyQueue.front());
            bodyQueue.pop_front();
        }
       
        return score;
    }
private:
    queue<pair<int, int>> foodQueue;
    int width, height;
    set<pair<int, int>> bodySet; // snake body set
    deque<pair<int, int>> bodyQueue; // snake body queue
    pair<int, int> currPos;
    int score;
};
相关推荐
糖葫芦君36 分钟前
Policy Gradient【强化学习的数学原理】
算法
向阳@向远方2 小时前
第二章 简单程序设计
开发语言·c++·算法
github_czy3 小时前
RRF (Reciprocal Rank Fusion) 排序算法详解
算法·排序算法
许愿与你永世安宁4 小时前
力扣343 整数拆分
数据结构·算法·leetcode
爱coding的橙子4 小时前
每日算法刷题Day42 7.5:leetcode前缀和3道题,用时2h
算法·leetcode·职场和发展
满分观察网友z4 小时前
从一次手滑,我洞悉了用户输入的所有可能性(3330. 找到初始输入字符串 I)
算法
YuTaoShao5 小时前
【LeetCode 热题 100】73. 矩阵置零——(解法二)空间复杂度 O(1)
java·算法·leetcode·矩阵
Heartoxx5 小时前
c语言-指针(数组)练习2
c语言·数据结构·算法
大熊背5 小时前
图像处理专业书籍以及网络资源总结
人工智能·算法·microsoft
满分观察网友z5 小时前
别怕树!一层一层剥开它的心:用BFS/DFS优雅计算层平均值(637. 二叉树的层平均值)
算法