mirror of
https://gitee.com/coder-xiaomo/leetcode-problemset
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186 lines
28 KiB
JSON
186 lines
28 KiB
JSON
{
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"questionId": "936",
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"questionFrontendId": "900",
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"categoryTitle": "Algorithms",
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"boundTopicId": 1240,
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"title": "RLE Iterator",
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"titleSlug": "rle-iterator",
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"content": "<p>We can use run-length encoding (i.e., <strong>RLE</strong>) to encode a sequence of integers. In a run-length encoded array of even length <code>encoding</code> (<strong>0-indexed</strong>), for all even <code>i</code>, <code>encoding[i]</code> tells us the number of times that the non-negative integer value <code>encoding[i + 1]</code> is repeated in the sequence.</p>\n\n<ul>\n\t<li>For example, the sequence <code>arr = [8,8,8,5,5]</code> can be encoded to be <code>encoding = [3,8,2,5]</code>. <code>encoding = [3,8,0,9,2,5]</code> and <code>encoding = [2,8,1,8,2,5]</code> are also valid <strong>RLE</strong> of <code>arr</code>.</li>\n</ul>\n\n<p>Given a run-length encoded array, design an iterator that iterates through it.</p>\n\n<p>Implement the <code>RLEIterator</code> class:</p>\n\n<ul>\n\t<li><code>RLEIterator(int[] encoded)</code> Initializes the object with the encoded array <code>encoded</code>.</li>\n\t<li><code>int next(int n)</code> Exhausts the next <code>n</code> elements and returns the last element exhausted in this way. If there is no element left to exhaust, return <code>-1</code> instead.</li>\n</ul>\n\n<p> </p>\n<p><strong class=\"example\">Example 1:</strong></p>\n\n<pre>\n<strong>Input</strong>\n["RLEIterator", "next", "next", "next", "next"]\n[[[3, 8, 0, 9, 2, 5]], [2], [1], [1], [2]]\n<strong>Output</strong>\n[null, 8, 8, 5, -1]\n\n<strong>Explanation</strong>\nRLEIterator rLEIterator = new RLEIterator([3, 8, 0, 9, 2, 5]); // This maps to the sequence [8,8,8,5,5].\nrLEIterator.next(2); // exhausts 2 terms of the sequence, returning 8. The remaining sequence is now [8, 5, 5].\nrLEIterator.next(1); // exhausts 1 term of the sequence, returning 8. The remaining sequence is now [5, 5].\nrLEIterator.next(1); // exhausts 1 term of the sequence, returning 5. The remaining sequence is now [5].\nrLEIterator.next(2); // exhausts 2 terms, returning -1. This is because the first term exhausted was 5,\nbut the second term did not exist. Since the last term exhausted does not exist, we return -1.\n</pre>\n\n<p> </p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>2 <= encoding.length <= 1000</code></li>\n\t<li><code>encoding.length</code> is even.</li>\n\t<li><code>0 <= encoding[i] <= 10<sup>9</sup></code></li>\n\t<li><code>1 <= n <= 10<sup>9</sup></code></li>\n\t<li>At most <code>1000</code> calls will be made to <code>next</code>.</li>\n</ul>\n",
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"translatedTitle": "RLE 迭代器",
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"translatedContent": "<p>我们可以使用游程编码(即 <strong>RLE </strong>)来编码一个整数序列。在偶数长度 <code>encoding</code> ( <strong>从 0 开始</strong> )的游程编码数组中，对于所有偶数 <code>i</code> ，<code>encoding[i]</code> 告诉我们非负整数 <code>encoding[i + 1]</code> 在序列中重复的次数。</p>\n\n<ul>\n\t<li>例如，序列 <code>arr = [8,8,8,5,5]</code> 可以被编码为 <code>encoding =[3,8,2,5]</code> 。<code>encoding =[3,8,0,9,2,5]</code> 和 <code>encoding =[2,8,1,8,2,5]</code> 也是 <code>arr</code> 有效的 <strong>RLE</strong> 。</li>\n</ul>\n\n<p>给定一个游程长度的编码数组，设计一个迭代器来遍历它。</p>\n\n<p>实现 <code>RLEIterator</code> 类:</p>\n\n<ul>\n\t<li><code>RLEIterator(int[] encoded)</code> 用编码后的数组初始化对象。</li>\n\t<li><code>int next(int n)</code> 以这种方式耗尽后 <code>n</code> 个元素并返回最后一个耗尽的元素。如果没有剩余的元素要耗尽，则返回 <code>-1</code> 。</li>\n</ul>\n\n<p> </p>\n\n<p><strong>示例 1：</strong></p>\n\n<pre>\n<strong>输入：\n</strong>[\"RLEIterator\",\"next\",\"next\",\"next\",\"next\"]\n[[[3,8,0,9,2,5]],[2],[1],[1],[2]]\n<strong>输出：\n</strong>[null,8,8,5,-1]\n<strong>解释：</strong>\nRLEIterator rLEIterator = new RLEIterator([3, 8, 0, 9, 2, 5]); // 这映射到序列 [8,8,8,5,5]。\nrLEIterator.next(2); // 耗去序列的 2 个项，返回 8。现在剩下的序列是 [8, 5, 5]。\nrLEIterator.next(1); // 耗去序列的 1 个项，返回 8。现在剩下的序列是 [5, 5]。\nrLEIterator.next(1); // 耗去序列的 1 个项，返回 5。现在剩下的序列是 [5]。\nrLEIterator.next(2); // 耗去序列的 2 个项，返回 -1。 这是由于第一个被耗去的项是 5，\n但第二个项并不存在。由于最后一个要耗去的项不存在，我们返回 -1。\n</pre>\n\n<p> </p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li><code>2 <= encoding.length <= 1000</code></li>\n\t<li><code>encoding.length</code> 为偶</li>\n\t<li><code>0 <= encoding[i] <= 10<sup>9</sup></code></li>\n\t<li><code>1 <= n <= 10<sup>9</sup></code></li>\n\t<li>每个测试用例调用<code>next </code>不高于 <code>1000</code> 次 </li>\n</ul>\n",
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"code": "class RLEIterator {\npublic:\n RLEIterator(vector<int>& encoding) {\n\n }\n \n int next(int n) {\n\n }\n};\n\n/**\n * Your RLEIterator object will be instantiated and called as such:\n * RLEIterator* obj = new RLEIterator(encoding);\n * int param_1 = obj->next(n);\n */",
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"code": "class RLEIterator {\n\n public RLEIterator(int[] encoding) {\n\n }\n \n public int next(int n) {\n\n }\n}\n\n/**\n * Your RLEIterator object will be instantiated and called as such:\n * RLEIterator obj = new RLEIterator(encoding);\n * int param_1 = obj.next(n);\n */",
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"code": "class RLEIterator(object):\n\n def __init__(self, encoding):\n \"\"\"\n :type encoding: List[int]\n \"\"\"\n\n\n def next(self, n):\n \"\"\"\n :type n: int\n :rtype: int\n \"\"\"\n\n\n\n# Your RLEIterator object will be instantiated and called as such:\n# obj = RLEIterator(encoding)\n# param_1 = obj.next(n)",
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"code": "class RLEIterator:\n\n def __init__(self, encoding: List[int]):\n\n\n def next(self, n: int) -> int:\n\n\n\n# Your RLEIterator object will be instantiated and called as such:\n# obj = RLEIterator(encoding)\n# param_1 = obj.next(n)",
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"code": "\n\n\ntypedef struct {\n \n} RLEIterator;\n\n\nRLEIterator* rLEIteratorCreate(int* encoding, int encodingSize) {\n \n}\n\nint rLEIteratorNext(RLEIterator* obj, int n) {\n \n}\n\nvoid rLEIteratorFree(RLEIterator* obj) {\n \n}\n\n/**\n * Your RLEIterator struct will be instantiated and called as such:\n * RLEIterator* obj = rLEIteratorCreate(encoding, encodingSize);\n * int param_1 = rLEIteratorNext(obj, n);\n \n * rLEIteratorFree(obj);\n*/",
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"code": "public class RLEIterator {\n\n public RLEIterator(int[] encoding) {\n\n }\n \n public int Next(int n) {\n\n }\n}\n\n/**\n * Your RLEIterator object will be instantiated and called as such:\n * RLEIterator obj = new RLEIterator(encoding);\n * int param_1 = obj.Next(n);\n */",
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"code": "/**\n * @param {number[]} encoding\n */\nvar RLEIterator = function(encoding) {\n\n};\n\n/** \n * @param {number} n\n * @return {number}\n */\nRLEIterator.prototype.next = function(n) {\n\n};\n\n/**\n * Your RLEIterator object will be instantiated and called as such:\n * var obj = new RLEIterator(encoding)\n * var param_1 = obj.next(n)\n */",
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"code": "class RLEIterator {\n constructor(encoding: number[]) {\n \n }\n\n next(n: number): number {\n \n }\n}\n\n/**\n * Your RLEIterator object will be instantiated and called as such:\n * var obj = new RLEIterator(encoding)\n * var param_1 = obj.next(n)\n */",
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"code": "class RLEIterator {\n /**\n * @param Integer[] $encoding\n */\n function __construct($encoding) {\n\n }\n\n /**\n * @param Integer $n\n * @return Integer\n */\n function next($n) {\n\n }\n}\n\n/**\n * Your RLEIterator object will be instantiated and called as such:\n * $obj = RLEIterator($encoding);\n * $ret_1 = $obj->next($n);\n */",
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"code": "\nclass RLEIterator {\n\n init(_ encoding: [Int]) {\n\n }\n \n func next(_ n: Int) -> Int {\n\n }\n}\n\n/**\n * Your RLEIterator object will be instantiated and called as such:\n * let obj = RLEIterator(encoding)\n * let ret_1: Int = obj.next(n)\n */",
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"code": "class RLEIterator(encoding: IntArray) {\n\n fun next(n: Int): Int {\n\n }\n\n}\n\n/**\n * Your RLEIterator object will be instantiated and called as such:\n * var obj = RLEIterator(encoding)\n * var param_1 = obj.next(n)\n */",
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"code": "class RLEIterator {\n\n RLEIterator(List<int> encoding) {\n \n }\n \n int next(int n) {\n \n }\n}\n\n/**\n * Your RLEIterator object will be instantiated and called as such:\n * RLEIterator obj = RLEIterator(encoding);\n * int param1 = obj.next(n);\n */",
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"code": "type RLEIterator struct {\n\n}\n\n\nfunc Constructor(encoding []int) RLEIterator {\n\n}\n\n\nfunc (this *RLEIterator) Next(n int) int {\n\n}\n\n\n/**\n * Your RLEIterator object will be instantiated and called as such:\n * obj := Constructor(encoding);\n * param_1 := obj.Next(n);\n */",
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"code": "class RLEIterator\n\n=begin\n :type encoding: Integer[]\n=end\n def initialize(encoding)\n\n end\n\n\n=begin\n :type n: Integer\n :rtype: Integer\n=end\n def next(n)\n\n end\n\n\nend\n\n# Your RLEIterator object will be instantiated and called as such:\n# obj = RLEIterator.new(encoding)\n# param_1 = obj.next(n)",
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