{ "data": { "question": { "questionId": "100171", "questionFrontendId": "面试题 04.01", "categoryTitle": "LCCI", "boundTopicId": 46021, "title": "Route Between Nodes LCCI ", "titleSlug": "route-between-nodes-lcci", "content": "

Given a directed graph, design an algorithm to find out whether there is a route between two nodes.

\r\n\r\n

Example1:

\r\n\r\n
\r\n Input: n = 3, graph = [[0, 1], [0, 2], [1, 2], [1, 2]], start = 0, target = 2\r\n Output: true\r\n
\r\n\r\n

Example2:

\r\n\r\n
\r\n Input: n = 5, graph = [[0, 1], [0, 2], [0, 4], [0, 4], [0, 1], [1, 3], [1, 4], [1, 3], [2, 3], [3, 4]], start = 0, target = 4\r\n Output true\r\n
\r\n\r\n

Note:

\r\n\r\n
    \r\n\t
  1. 0 <= n <= 100000
    2. \r\n\t
  2. All node numbers are within the range [0, n].
    3. \r\n\t
  3. There might be self cycles and duplicated edges.
    4. \r\n
\r\n", "translatedTitle": "节点间通路", "translatedContent": "

节点间通路。给定有向图，设计一个算法，找出两个节点之间是否存在一条路径。

\n\n

示例 1：

\n\n
\n 输入：n = 3, graph = [[0, 1], [0, 2], [1, 2], [1, 2]], start = 0, target = 2\n 输出：true\n
\n\n

示例 2：

\n\n
\n 输入：n = 5, graph = [[0, 1], [0, 2], [0, 4], [0, 4], [0, 1], [1, 3], [1, 4], [1, 3], [2, 3], [3, 4]], start = 0, target = 4\n 输出：true\n
\n\n

提示：

\n\n
    \n\t
  1. 节点数量n在[0, 1e5]范围内。
    2. \n\t
  2. 节点编号大于等于 0 小于 n。
    3. \n\t
  3. 图中可能存在自环和平行边。
    4. \n
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(listof (listof exact-integer?)) exact-integer? exact-integer? boolean?)\n )", "__typename": "CodeSnippetNode" }, { "lang": "Erlang", "langSlug": "erlang", "code": "-spec find_whether_exists_path(N :: integer(), Graph :: [[integer()]], Start :: integer(), Target :: integer()) -> boolean().\nfind_whether_exists_path(N, Graph, Start, Target) ->\n .", "__typename": "CodeSnippetNode" }, { "lang": "Elixir", "langSlug": "elixir", "code": "defmodule Solution do\n @spec find_whether_exists_path(n :: integer, graph :: [[integer]], start :: integer, target :: integer) :: boolean\n def find_whether_exists_path(n, graph, start, target) do\n \n end\nend", "__typename": "CodeSnippetNode" }, { "lang": "Cangjie", "langSlug": "cangjie", "code": "class Solution {\n func findWhetherExistsPath(n: Int64, graph: Array>, start: Int64, target: Int64): Bool {\n\n }\n}", "__typename": "CodeSnippetNode" } ], "stats": "{\"totalAccepted\": \"35.7K\", \"totalSubmission\": \"67.7K\", \"totalAcceptedRaw\": 35702, \"totalSubmissionRaw\": 67675, \"acRate\": \"52.8%\"}", "hints": [ "有两个众所周知的算法可以做到这一点。其利弊是什么？" ], "solution": null, "status": null, "sampleTestCase": "3\n[[0, 1], [0, 2], [1, 2], [1, 2]]\n0\n2", "metaData": "{\"name\": \"findWhetherExistsPath\", \"params\": [{\"name\": \"n\", \"type\": \"integer\"}, {\"name\": \"graph\", \"type\": \"integer[][]\"}, {\"name\": \"start\", \"type\": \"integer\"}, {\"name\": \"target\", \"type\": \"integer\"}], \"return\": {\"type\": \"boolean\"}}", "judgerAvailable": true, "judgeType": "large", "mysqlSchemas": [], "enableRunCode": true, "envInfo": "{\"cpp\":[\"C++\",\"

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            "book": {
                "id": "2",
                "bookName": "《程序员面试金典（第 6 版）》",
                "pressName": "图灵公司",
                "source": "《程序员面试金典（第 6 版）》独家授权",
                "shortDescription": "英文版：Cracking the Coding Interview，源自硅谷程序员的面试红宝书",
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