{ "data": { "question": { "questionId": "3533", "questionFrontendId": "3248", "categoryTitle": "Algorithms", "boundTopicId": 2874939, "title": "Snake in Matrix", "titleSlug": "snake-in-matrix", "content": "
There is a snake in an n x n
matrix grid
and can move in four possible directions. Each cell in the grid
is identified by the position: grid[i][j] = (i * n) + j
.
The snake starts at cell 0 and follows a sequence of commands.
\n\nYou are given an integer n
representing the size of the grid
and an array of strings commands
where each command[i]
is either "UP"
, "RIGHT"
, "DOWN"
, and "LEFT"
. It's guaranteed that the snake will remain within the grid
boundaries throughout its movement.
Return the position of the final cell where the snake ends up after executing commands
.
\n
Example 1:
\n\nInput: n = 2, commands = ["RIGHT","DOWN"]
\n\nOutput: 3
\n\nExplanation:
\n\n0 | \n\t\t\t1 | \n\t\t
2 | \n\t\t\t3 | \n\t\t
0 | \n\t\t\t1 | \n\t\t
2 | \n\t\t\t3 | \n\t\t
0 | \n\t\t\t1 | \n\t\t
2 | \n\t\t\t3 | \n\t\t
Example 2:
\n\nInput: n = 3, commands = ["DOWN","RIGHT","UP"]
\n\nOutput: 1
\n\nExplanation:
\n\n0 | \n\t\t\t1 | \n\t\t\t2 | \n\t\t
3 | \n\t\t\t4 | \n\t\t\t5 | \n\t\t
6 | \n\t\t\t7 | \n\t\t\t8 | \n\t\t
0 | \n\t\t\t1 | \n\t\t\t2 | \n\t\t
3 | \n\t\t\t4 | \n\t\t\t5 | \n\t\t
6 | \n\t\t\t7 | \n\t\t\t8 | \n\t\t
0 | \n\t\t\t1 | \n\t\t\t2 | \n\t\t
3 | \n\t\t\t4 | \n\t\t\t5 | \n\t\t
6 | \n\t\t\t7 | \n\t\t\t8 | \n\t\t
0 | \n\t\t\t1 | \n\t\t\t2 | \n\t\t
3 | \n\t\t\t4 | \n\t\t\t5 | \n\t\t
6 | \n\t\t\t7 | \n\t\t\t8 | \n\t\t
\n
Constraints:
\n\n2 <= n <= 10
1 <= commands.length <= 100
commands
consists only of "UP"
, "RIGHT"
, "DOWN"
, and "LEFT"
.大小为 n x n
的矩阵 grid
中有一条蛇。蛇可以朝 四个可能的方向 移动。矩阵中的每个单元格都使用位置进行标识: grid[i][j] = (i * n) + j
。
蛇从单元格 0 开始,并遵循一系列命令移动。
\n\n给你一个整数 n
表示 grid
的大小,另给你一个字符串数组 commands
,其中包括 \"UP\"
、\"RIGHT\"
、\"DOWN\"
和 \"LEFT\"
。题目测评数据保证蛇在整个移动过程中将始终位于 grid
边界内。
返回执行 commands
后蛇所停留的最终单元格的位置。
\n\n
示例 1:
\n\n输入:n = 2, commands = [\"RIGHT\",\"DOWN\"]
\n\n输出:3
\n\n解释:
\n\n0 | \n\t\t\t1 | \n\t\t
2 | \n\t\t\t3 | \n\t\t
0 | \n\t\t\t1 | \n\t\t
2 | \n\t\t\t3 | \n\t\t
0 | \n\t\t\t1 | \n\t\t
2 | \n\t\t\t3 | \n\t\t
示例 2:
\n\n输入:n = 3, commands = [\"DOWN\",\"RIGHT\",\"UP\"]
\n\n输出:1
\n\n解释:
\n\n0 | \n\t\t\t1 | \n\t\t\t2 | \n\t\t
3 | \n\t\t\t4 | \n\t\t\t5 | \n\t\t
6 | \n\t\t\t7 | \n\t\t\t8 | \n\t\t
0 | \n\t\t\t1 | \n\t\t\t2 | \n\t\t
3 | \n\t\t\t4 | \n\t\t\t5 | \n\t\t
6 | \n\t\t\t7 | \n\t\t\t8 | \n\t\t
0 | \n\t\t\t1 | \n\t\t\t2 | \n\t\t
3 | \n\t\t\t4 | \n\t\t\t5 | \n\t\t
6 | \n\t\t\t7 | \n\t\t\t8 | \n\t\t
0 | \n\t\t\t1 | \n\t\t\t2 | \n\t\t
3 | \n\t\t\t4 | \n\t\t\t5 | \n\t\t
6 | \n\t\t\t7 | \n\t\t\t8 | \n\t\t
\n\n
提示:
\n\n2 <= n <= 10
1 <= commands.length <= 100
commands
仅由 \"UP\"
、\"RIGHT\"
、\"DOWN\"
和 \"LEFT\"
组成。\\u7248\\u672c\\uff1a \\u7f16\\u8bd1\\u65f6\\uff0c\\u5c06\\u4f1a\\u91c7\\u7528 \\u4e3a\\u4e86\\u4f7f\\u7528\\u65b9\\u4fbf\\uff0c\\u5927\\u90e8\\u5206\\u6807\\u51c6\\u5e93\\u7684\\u5934\\u6587\\u4ef6\\u5df2\\u7ecf\\u88ab\\u81ea\\u52a8\\u5bfc\\u5165\\u3002<\\/p>\"],\"java\":[\"Java\",\" \\u7248\\u672c\\uff1a \\u4e3a\\u4e86\\u65b9\\u4fbf\\u8d77\\u89c1\\uff0c\\u5927\\u90e8\\u5206\\u6807\\u51c6\\u5e93\\u7684\\u5934\\u6587\\u4ef6\\u5df2\\u88ab\\u5bfc\\u5165\\u3002<\\/p>\\r\\n\\r\\n \\u5305\\u542b Pair \\u7c7b: https:\\/\\/docs.oracle.com\\/javase\\/8\\/javafx\\/api\\/javafx\\/util\\/Pair.html <\\/p>\"],\"python\":[\"Python\",\" \\u7248\\u672c\\uff1a \\u4e3a\\u4e86\\u65b9\\u4fbf\\u8d77\\u89c1\\uff0c\\u5927\\u90e8\\u5206\\u5e38\\u7528\\u5e93\\u5df2\\u7ecf\\u88ab\\u81ea\\u52a8 \\u5bfc\\u5165\\uff0c\\u5982\\uff1aarray<\\/a>, bisect<\\/a>, 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"editorType": "CKEDITOR",
"ugcQuestionId": null,
"style": "LEETCODE",
"exampleTestcases": "2\n[\"RIGHT\",\"DOWN\"]\n3\n[\"DOWN\",\"RIGHT\",\"UP\"]",
"__typename": "QuestionNode"
}
}
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out-of-bounds<\\/code> \\u548c
use-after-free<\\/code> \\u9519\\u8bef\\u3002<\\/p>\\r\\n\\r\\n
\\r\\nstruct hash_entry {\\r\\n int id; \\/* we'll use this field as the key *\\/\\r\\n char name[10];\\r\\n UT_hash_handle hh; \\/* makes this structure hashable *\\/\\r\\n};\\r\\n\\r\\nstruct hash_entry *users = NULL;\\r\\n\\r\\nvoid add_user(struct hash_entry *s) {\\r\\n HASH_ADD_INT(users, id, s);\\r\\n}\\r\\n<\\/pre>\\r\\n<\\/p>\\r\\n\\r\\n
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