leetcode-problemset/leetcode-cn/originData/transformed-array.json

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        "question": {
            "questionId": "3651",
            "questionFrontendId": "3379",
            "categoryTitle": "Algorithms",
            "boundTopicId": 3009890,
            "title": "Transformed Array",
            "titleSlug": "transformed-array",
            "content": "<p>You are given an integer array <code>nums</code> that represents a circular array. Your task is to create a new array <code>result</code> of the <strong>same</strong> size, following these rules:</p>\nFor each index <code>i</code> (where <code>0 &lt;= i &lt; nums.length</code>), perform the following <strong>independent</strong> actions:\n\n<ul>\n\t<li>If <code>nums[i] &gt; 0</code>: Start at index <code>i</code> and move <code>nums[i]</code> steps to the <strong>right</strong> in the circular array. Set <code>result[i]</code> to the value of the index where you land.</li>\n\t<li>If <code>nums[i] &lt; 0</code>: Start at index <code>i</code> and move <code>abs(nums[i])</code> steps to the <strong>left</strong> in the circular array. Set <code>result[i]</code> to the value of the index where you land.</li>\n\t<li>If <code>nums[i] == 0</code>: Set <code>result[i]</code> to <code>nums[i]</code>.</li>\n</ul>\n\n<p>Return the new array <code>result</code>.</p>\n\n<p><strong>Note:</strong> Since <code>nums</code> is circular, moving past the last element wraps around to the beginning, and moving before the first element wraps back to the end.</p>\n\n<p>&nbsp;</p>\n<p><strong class=\"example\">Example 1:</strong></p>\n\n<div class=\"example-block\">\n<p><strong>Input:</strong> <span class=\"example-io\">nums = [3,-2,1,1]</span></p>\n\n<p><strong>Output:</strong> <span class=\"example-io\">[1,1,1,3]</span></p>\n\n<p><strong>Explanation:</strong></p>\n\n<ul>\n\t<li>For <code>nums[0]</code> that is equal to 3, If we move 3 steps to right, we reach <code>nums[3]</code>. So <code>result[0]</code> should be 1.</li>\n\t<li>For <code>nums[1]</code> that is equal to -2, If we move 2 steps to left, we reach <code>nums[3]</code>. So <code>result[1]</code> should be 1.</li>\n\t<li>For <code>nums[2]</code> that is equal to 1, If we move 1 step to right, we reach <code>nums[3]</code>. So <code>result[2]</code> should be 1.</li>\n\t<li>For <code>nums[3]</code> that is equal to 1, If we move 1 step to right, we reach <code>nums[0]</code>. So <code>result[3]</code> should be 3.</li>\n</ul>\n</div>\n\n<p><strong class=\"example\">Example 2:</strong></p>\n\n<div class=\"example-block\">\n<p><strong>Input:</strong> <span class=\"example-io\">nums = [-1,4,-1]</span></p>\n\n<p><strong>Output:</strong> <span class=\"example-io\">[-1,-1,4]</span></p>\n\n<p><strong>Explanation:</strong></p>\n\n<ul>\n\t<li>For <code>nums[0]</code> that is equal to -1, If we move 1 step to left, we reach <code>nums[2]</code>. So <code>result[0]</code> should be -1.</li>\n\t<li>For <code>nums[1]</code> that is equal to 4, If we move 4 steps to right, we reach <code>nums[2]</code>. So <code>result[1]</code> should be -1.</li>\n\t<li>For <code>nums[2]</code> that is equal to -1, If we move 1 step to left, we reach <code>nums[1]</code>. So <code>result[2]</code> should be 4.</li>\n</ul>\n</div>\n\n<p>&nbsp;</p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= nums.length &lt;= 100</code></li>\n\t<li><code>-100 &lt;= nums[i] &lt;= 100</code></li>\n</ul>\n",
            "translatedTitle": "转换数组",
            "translatedContent": "<p>给你一个整数数组 <code>nums</code>，它表示一个循环数组。请你遵循以下规则创建一个大小&nbsp;<strong>相同&nbsp;</strong>的新数组 <code>result</code>&nbsp;：</p>\n对于每个下标&nbsp;<code>i</code>（其中 <code>0 &lt;= i &lt; nums.length</code>），独立执行以下操作：\n\n<ul>\n\t<li>如果 <code>nums[i] &gt; 0</code>：从下标&nbsp;<code>i</code> 开始，向&nbsp;<strong>右&nbsp;</strong>移动 <code>nums[i]</code> 步，在循环数组中落脚的下标对应的值赋给 <code>result[i]</code>。</li>\n\t<li>如果 <code>nums[i] &lt; 0</code>：从下标&nbsp;<code>i</code> 开始，向&nbsp;<strong>左&nbsp;</strong>移动 <code>abs(nums[i])</code> 步，在循环数组中落脚的下标对应的值赋给 <code>result[i]</code>。</li>\n\t<li>如果 <code>nums[i] == 0</code>：将 <code>nums[i]</code> 的值赋给 <code>result[i]</code>。</li>\n</ul>\n\n<p>返回新数组 <code>result</code>。</p>\n\n<p><strong>注意：</strong>由于 <code>nums</code> 是循环数组，向右移动超过最后一个元素时将回到开头，向左移动超过第一个元素时将回到末尾。</p>\n\n<p>&nbsp;</p>\n\n<p><strong class=\"example\">示例 1：</strong></p>\n\n<div class=\"example-block\">\n<p><strong>输入：</strong> <span class=\"example-io\">nums = [3,-2,1,1]</span></p>\n\n<p><strong>输出：</strong> <span class=\"example-io\">[1,1,1,3]</span></p>\n\n<p><strong>解释：</strong></p>\n\n<ul>\n\t<li>对于 <code>nums[0]</code> 等于 3，向右移动 3 步到 <code>nums[3]</code>，因此 <code>result[0]</code> 为 1。</li>\n\t<li>对于 <code>nums[1]</code> 等于 -2，向左移动 2 步到 <code>nums[3]</code>，因此 <code>result[1]</code> 为 1。</li>\n\t<li>对于 <code>nums[2]</code> 等于 1，向右移动 1 步到 <code>nums[3]</code>，因此 <code>result[2]</code> 为 1。</li>\n\t<li>对于 <code>nums[3]</code> 等于 1，向右移动 1 步到 <code>nums[0]</code>，因此 <code>result[3]</code> 为 3。</li>\n</ul>\n</div>\n\n<p><strong class=\"example\">示例 2：</strong></p>\n\n<div class=\"example-block\">\n<p><strong>输入：</strong> <span class=\"example-io\">nums = [-1,4,-1]</span></p>\n\n<p><strong>输出：</strong> <span class=\"example-io\">[-1,-1,4]</span></p>\n\n<p><strong>解释：</strong></p>\n\n<ul>\n\t<li>对于 <code>nums[0]</code> 等于 -1，向左移动 1 步到 <code>nums[2]</code>，因此 <code>result[0]</code> 为 -1。</li>\n\t<li>对于 <code>nums[1]</code> 等于 4，向右移动 4 步到 <code>nums[2]</code>，因此 <code>result[1]</code> 为 -1。</li>\n\t<li>对于 <code>nums[2]</code> 等于 -1，向左移动 1 步到 <code>nums[1]</code>，因此 <code>result[2]</code> 为 4。</li>\n</ul>\n</div>\n\n<p>&nbsp;</p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= nums.length &lt;= 100</code></li>\n\t<li><code>-100 &lt;= nums[i] &lt;= 100</code></li>\n</ul>\n",
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target=\\\"_blank\\\">sortedcontainers<\\/a> \\u5e93\\u3002<\\/p>\"],\"scala\":[\"Scala\",\"<p>\\u7248\\u672c\\uff1a<code>Scala 3.3.1<\\/code><\\/p>\"],\"kotlin\":[\"Kotlin\",\"<p>\\u7248\\u672c\\uff1a<code>Kotlin 1.9.0<\\/code><\\/p>\\r\\n\\r\\n<p>\\u6211\\u4eec\\u4f7f\\u7528\\u7684\\u662f JetBrains \\u63d0\\u4f9b\\u7684 experimental compiler\\u3002\\u5982\\u679c\\u60a8\\u8ba4\\u4e3a\\u60a8\\u9047\\u5230\\u4e86\\u7f16\\u8bd1\\u5668\\u76f8\\u5173\\u7684\\u95ee\\u9898\\uff0c\\u8bf7\\u5411\\u6211\\u4eec\\u53cd\\u9988<\\/p>\"],\"rust\":[\"Rust\",\"<p>\\u7248\\u672c\\uff1a<code>rust 1.79.0<\\/code><\\/p>\\r\\n\\r\\n<p>\\u652f\\u6301 crates.io \\u7684 <a href=\\\"https:\\/\\/crates.io\\/crates\\/rand\\\" target=\\\"_blank\\\">rand<\\/a> \\u548c <a href=\\\"https:\\/\\/crates.io\\/crates\\/regex\\\" target=\\\"_blank\\\">regex<\\/a><\\/p>\"],\"php\":[\"PHP\",\"<p><code>PHP 8.2<\\/code>.<\\/p>\\r\\n\\r\\n<p>With bcmath module.<\\/p>\"],\"typescript\":[\"TypeScript\",\"<p>TypeScript 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href=\\\"https:\\/\\/docs.racket-lang.org\\/guide\\/performance.html#%28tech._c%29\\\" target=\\\"_blank\\\">Racket CS<\\/a> v8.15<\\/p>\\r\\n\\r\\n<p>\\u4f7f\\u7528 #lang racket<\\/p>\\r\\n\\r\\n<p>\\u5df2\\u9884\\u5148 (require data\\/gvector data\\/queue data\\/order data\\/heap). \\u82e5\\u9700\\u4f7f\\u7528\\u5176\\u5b83\\u6570\\u636e\\u7ed3\\u6784\\uff0c\\u53ef\\u81ea\\u884c require\\u3002<\\/p>\"],\"erlang\":[\"Erlang\",\"Erlang\\/OTP 26\"],\"elixir\":[\"Elixir\",\"Elixir 1.17 with Erlang\\/OTP 26\"],\"dart\":[\"Dart\",\"<p>Dart 3.2\\u3002\\u60a8\\u53ef\\u4ee5\\u4f7f\\u7528 <a href=\\\"https:\\/\\/pub.dev\\/packages\\/collection\\/versions\\/1.18.0\\\" target=\\\"_blank\\\">collection<\\/a> \\u5305<\\/p>\\r\\n\\r\\n<p>\\u60a8\\u7684\\u4ee3\\u7801\\u5c06\\u4f1a\\u88ab\\u4e0d\\u7f16\\u8bd1\\u76f4\\u63a5\\u8fd0\\u884c<\\/p>\"],\"cangjie\":[\"Cangjie\",\"<p>\\u7248\\u672c\\uff1a0.53.11 (cjnative)<\\/p>\\r\\n\\r\\n<p>\\u7f16\\u8bd1\\u53c2\\u6570\\uff1a<code>-O2 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