{ "data": { "question": { "questionId": "2239", "questionFrontendId": "2120", "categoryTitle": "Algorithms", "boundTopicId": 1174200, "title": "Execution of All Suffix Instructions Staying in a Grid", "titleSlug": "execution-of-all-suffix-instructions-staying-in-a-grid", "content": "
There is an n x n
grid, with the top-left cell at (0, 0)
and the bottom-right cell at (n - 1, n - 1)
. You are given the integer n
and an integer array startPos
where startPos = [startrow, startcol]
indicates that a robot is initially at cell (startrow, startcol)
.
You are also given a 0-indexed string s
of length m
where s[i]
is the ith
instruction for the robot: 'L'
(move left), 'R'
(move right), 'U'
(move up), and 'D'
(move down).
The robot can begin executing from any ith
instruction in s
. It executes the instructions one by one towards the end of s
but it stops if either of these conditions is met:
Return an array answer
of length m
where answer[i]
is the number of instructions the robot can execute if the robot begins executing from the ith
instruction in s
.
\n
Example 1:
\n\n\nInput: n = 3, startPos = [0,1], s = "RRDDLU"\nOutput: [1,5,4,3,1,0]\nExplanation: Starting from startPos and beginning execution from the ith instruction:\n- 0th: "RRDDLU". Only one instruction "R" can be executed before it moves off the grid.\n- 1st: "RDDLU". All five instructions can be executed while it stays in the grid and ends at (1, 1).\n- 2nd: "DDLU". All four instructions can be executed while it stays in the grid and ends at (1, 0).\n- 3rd: "DLU". All three instructions can be executed while it stays in the grid and ends at (0, 0).\n- 4th: "LU". Only one instruction "L" can be executed before it moves off the grid.\n- 5th: "U". If moving up, it would move off the grid.\n\n\n
Example 2:
\n\n\nInput: n = 2, startPos = [1,1], s = "LURD"\nOutput: [4,1,0,0]\nExplanation:\n- 0th: "LURD".\n- 1st: "URD".\n- 2nd: "RD".\n- 3rd: "D".\n\n\n
Example 3:
\n\n\nInput: n = 1, startPos = [0,0], s = "LRUD"\nOutput: [0,0,0,0]\nExplanation: No matter which instruction the robot begins execution from, it would move off the grid.\n\n\n
\n
Constraints:
\n\nm == s.length
1 <= n, m <= 500
startPos.length == 2
0 <= startrow, startcol < n
s
consists of 'L'
, 'R'
, 'U'
, and 'D'
.现有一个 n x n
大小的网格,左上角单元格坐标 (0, 0)
,右下角单元格坐标 (n - 1, n - 1)
。给你整数 n
和一个整数数组 startPos
,其中 startPos = [startrow, startcol]
表示机器人最开始在坐标为 (startrow, startcol)
的单元格上。
另给你一个长度为 m
、下标从 0 开始的字符串 s
,其中 s[i]
是对机器人的第 i
条指令:'L'
(向左移动),'R'
(向右移动),'U'
(向上移动)和 'D'
(向下移动)。
机器人可以从 s
中的任一第 i
条指令开始执行。它将会逐条执行指令直到 s
的末尾,但在满足下述条件之一时,机器人将会停止:
返回一个长度为 m
的数组 answer
,其中 answer[i]
是机器人从第 i
条指令 开始 ,可以执行的 指令数目 。
\n\n
示例 1:
\n\n\n\n\n输入:n = 3, startPos = [0,1], s = \"RRDDLU\"\n输出:[1,5,4,3,1,0]\n解释:机器人从 startPos 出发,并从第 i 条指令开始执行:\n- 0: \"RRDDLU\" 在移动到网格外之前,只能执行一条 \"R\" 指令。\n- 1: \"RDDLU\" 可以执行全部五条指令,机器人仍在网格内,最终到达 (0, 0) 。\n- 2: \"DDLU\" 可以执行全部四条指令,机器人仍在网格内,最终到达 (0, 0) 。\n- 3: \"DLU\" 可以执行全部三条指令,机器人仍在网格内,最终到达 (0, 0) 。\n- 4: \"LU\" 在移动到网格外之前,只能执行一条 \"L\" 指令。\n- 5: \"U\" 如果向上移动,将会移动到网格外。\n\n\n
示例 2:
\n\n\n\n\n输入:n = 2, startPos = [1,1], s = \"LURD\"\n输出:[4,1,0,0]\n解释:\n- 0: \"LURD\"\n- 1: \"URD\"\n- 2: \"RD\"\n- 3: \"D\"\n\n\n
示例 3:
\n\n\n\n\n输入:n = 1, startPos = [0,0], s = \"LRUD\"\n输出:[0,0,0,0]\n解释:无论机器人从哪条指令开始执行,都会移动到网格外。\n\n\n
\n\n
提示:
\n\nm == s.length
1 <= n, m <= 500
startPos.length == 2
0 <= startrow, startcol < n
s
由 'L'
、'R'
、'U'
和 'D'
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out-of-bounds<\\/code>\\u548c
use-after-free<\\/code>\\u9519\\u8bef\\u3002<\\/p>\\r\\n\\r\\n
OpenJDK 17<\\/code>\\u3002\\u53ef\\u4ee5\\u4f7f\\u7528Java 8\\u7684\\u7279\\u6027\\u4f8b\\u5982\\uff0clambda expressions \\u548c stream API\\u3002<\\/p>\\r\\n\\r\\n
Python 2.7.12<\\/code><\\/p>\\r\\n\\r\\n
GCC 8.2<\\/code>\\uff0c\\u91c7\\u7528GNU99\\u6807\\u51c6\\u3002<\\/p>\\r\\n\\r\\n
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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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Node.js 16.13.2<\\/code><\\/p>\\r\\n\\r\\n
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