mirror of
https://gitee.com/coder-xiaomo/leetcode-problemset
synced 2025-01-10 18:48:13 +08:00
161 lines
20 KiB
JSON
161 lines
20 KiB
JSON
{
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"data": {
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"question": {
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"questionId": "1000568",
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"questionFrontendId": "LCP 80",
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"categoryTitle": "Algorithms",
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"boundTopicId": 2258736,
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"title": "生物进化录",
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"titleSlug": "qoQAMX",
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"content": "English description is not available for the problem. Please switch to Chinese.",
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"translatedTitle": "生物进化录",
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"translatedContent": "在永恒之森中,存在着一本生物进化录,以 **一个树形结构** 记载了所有生物的演化过程。经过观察并整理了各节点间的关系,`parents[i]` 表示编号 `i` 节点的父节点编号(根节点的父节点为 `-1`)。\n\n为了探索和记录其中的演化规律,队伍中的炼金术师提出了一种方法,可以以字符串的形式将其复刻下来,规则如下:\n- 初始只有一个根节点,表示演化的起点,依次记录 `01` 字符串中的字符,\n- 如果记录 `0`,则在当前节点下添加一个子节点,并将指针指向新添加的子节点;\n- 如果记录 `1`,则将指针回退到当前节点的父节点处。\n\n现在需要应用上述的记录方法,复刻下它的演化过程。请返回能够复刻演化过程的字符串中, **字典序最小** 的 `01` 字符串。\n\n**注意:**\n- 节点指针最终可以停在任何节点上,不一定要回到根节点。\n\n**示例 1:**\n> 输入:`parents = [-1,0,0,2]`\n>\n> 输出:`\"00110\"`\n>\n>解释:树结构如下图所示,共存在 2 种记录方案:\n>第 1 种方案为:0(记录编号 1 的节点) -> 1(回退至节点 0) -> 0(记录编号 2 的节点) -> 0((记录编号 3 的节点))\n>第 2 种方案为:0(记录编号 2 的节点) -> 0(记录编号 3 的节点) -> 1(回退至节点 2) -> 1(回退至节点 0) -> 0(记录编号 1 的节点)\n>返回字典序更小的 `\"00110\"`\n![image.png](https://pic.leetcode.cn/1682319485-cRVudI-image.png){:width=120px}![进化 (3).gif](https://pic.leetcode.cn/1682412701-waHdnm-%E8%BF%9B%E5%8C%96%20\\(3\\).gif){:width=320px}\n\n\n\n**示例 2:**\n> 输入:`parents = [-1,0,0,1,2,2]`\n>\n> 输出:`\"00101100\"`\n\n**提示:**\n\n- `1 <= parents.length <= 10^4`\n- `-1 <= parents[i] < i` (即父节点编号小于子节点)",
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"difficulty": "Hard",
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"likes": 4,
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"lang": "C++",
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"langSlug": "cpp",
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"code": "class Solution {\npublic:\n string evolutionaryRecord(vector<int>& parents) {\n\n }\n};",
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"code": "class Solution {\n public String evolutionaryRecord(int[] parents) {\n\n }\n}",
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"code": "class Solution(object):\n def evolutionaryRecord(self, parents):\n \"\"\"\n :type parents: List[int]\n :rtype: str\n \"\"\"",
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"code": "char* evolutionaryRecord(int* parents, int parentsSize){\n\n}",
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"code": "/**\n * @param {number[]} parents\n * @return {string}\n */\nvar evolutionaryRecord = function(parents) {\n\n};",
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"code": "# @param {Integer[]} parents\n# @return {String}\ndef evolutionary_record(parents)\n\nend",
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"code": "object Solution {\n def evolutionaryRecord(parents: Array[Int]): String = {\n\n }\n}",
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"code": "impl Solution {\n pub fn evolutionary_record(parents: Vec<i32>) -> String {\n\n }\n}",
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