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https://gitee.com/coder-xiaomo/leetcode-problemset
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59 lines
8.8 KiB
JSON
59 lines
8.8 KiB
JSON
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"questionFrontendId": "2695",
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"categoryTitle": "JavaScript",
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"boundTopicId": 2275511,
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"title": "Array Wrapper",
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"titleSlug": "array-wrapper",
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"content": "<p>Create a class <code>ArrayWrapper</code> that accepts an array of integers in its constructor. This class should have two features:</p>\n\n<ul>\n\t<li>When two instances of this class are added together with the <code>+</code> operator, the resulting value is the sum of all the elements in both arrays.</li>\n\t<li>When the <code>String()</code> function is called on the instance, it will return a comma separated string surrounded by brackets. For example, <code>[1,2,3]</code>.</li>\n</ul>\n\n<p> </p>\n<p><strong class=\"example\">Example 1:</strong></p>\n\n<pre>\n<strong>Input:</strong> nums = [[1,2],[3,4]], operation = "Add"\n<strong>Output:</strong> 10\n<strong>Explanation:</strong>\nconst obj1 = new ArrayWrapper([1,2]);\nconst obj2 = new ArrayWrapper([3,4]);\nobj1 + obj2; // 10\n</pre>\n\n<p><strong class=\"example\">Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> nums = [[23,98,42,70]], operation = "String"\n<strong>Output:</strong> "[23,98,42,70]"\n<strong>Explanation:</strong>\nconst obj = new ArrayWrapper([23,98,42,70]);\nString(obj); // "[23,98,42,70]"\n</pre>\n\n<p><strong class=\"example\">Example 3:</strong></p>\n\n<pre>\n<strong>Input:</strong> nums = [[],[]], operation = "Add"\n<strong>Output:</strong> 0\n<strong>Explanation:</strong>\nconst obj1 = new ArrayWrapper([]);\nconst obj2 = new ArrayWrapper([]);\nobj1 + obj2; // 0\n</pre>\n\n<p> </p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>0 <= nums.length <= 1000</code></li>\n\t<li><code>0 <= nums[i] <= 1000</code></li>\n\t<li><code>Note: nums is the array passed to the constructor</code></li>\n</ul>\n",
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"translatedTitle": "包装数组",
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"translatedContent": "<p>创建一个名为 <code>ArrayWrapper</code> 的类,它在其构造函数中接受一个整数数组作为参数。该类应具有以下两个特性:</p>\n\n<ul>\n\t<li>当使用 <code>+</code> 运算符将两个该类的实例相加时,结果值为两个数组中所有元素的总和。</li>\n\t<li>当在实例上调用 <code>String()</code> 函数时,它将返回一个由逗号分隔的括在方括号中的字符串。例如,<code>[1,2,3]</code> 。</li>\n</ul>\n\n<p> </p>\n\n<p><strong class=\"example\">示例 1:</strong></p>\n\n<pre>\n<b>输入:</b>nums = [[1,2],[3,4]], operation = \"Add\"\n<b>输出:</b>10\n<b>解释:</b>\nconst obj1 = new ArrayWrapper([1,2]);\nconst obj2 = new ArrayWrapper([3,4]);\nobj1 + obj2; // 10\n</pre>\n\n<p><strong class=\"example\">示例 2:</strong></p>\n\n<pre>\n<b>输入:</b>nums = [[23,98,42,70]], operation = \"String\"\n<b>输出:</b>\"[23,98,42,70]\"\n<strong>解释:</strong>\nconst obj = new ArrayWrapper([23,98,42,70]);\nString(obj); // \"[23,98,42,70]\"\n</pre>\n\n<p><strong class=\"example\">示例 3:</strong></p>\n\n<pre>\n<b>输入:</b>nums = [[],[]], operation = \"Add\"\n<b>输出:</b>0\n<strong>解释:</strong>\nconst obj1 = new ArrayWrapper([]);\nconst obj2 = new ArrayWrapper([]);\nobj1 + obj2; // 0\n</pre>\n\n<p> </p>\n\n<p><b>提示:</b></p>\n\n<ul>\n\t<li><code>0 <= nums.length <= 1000</code></li>\n\t<li><code>0 <= nums[i] <= 1000</code></li>\n\t<li><code>注意:nums 是传递给构造函数的数组。</code></li>\n</ul>\n",
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"difficulty": "Easy",
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"code": "/**\n * @param {number[]} nums\n * @return {void}\n */\nvar ArrayWrapper = function(nums) {\n \n};\n\n/**\n * @return {number}\n */\nArrayWrapper.prototype.valueOf = function() {\n \n}\n\n/**\n * @return {string}\n */\nArrayWrapper.prototype.toString = function() {\n \n}\n\n/**\n * const obj1 = new ArrayWrapper([1,2]);\n * const obj2 = new ArrayWrapper([3,4]);\n * obj1 + obj2; // 10\n * String(obj1); // \"[1,2]\"\n * String(obj2); // \"[3,4]\"\n */",
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