mirror of
https://gitee.com/coder-xiaomo/leetcode-problemset
synced 2025-01-11 02:58:13 +08:00
180 lines
19 KiB
JSON
180 lines
19 KiB
JSON
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"boundTopicId": 1690,
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"title": "Count Numbers with Unique Digits",
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"content": "<p>Given an integer <code>n</code>, return the count of all numbers with unique digits, <code>x</code>, where <code>0 <= x < 10<sup>n</sup></code>.</p>\n\n<p> </p>\n<p><strong>Example 1:</strong></p>\n\n<pre>\n<strong>Input:</strong> n = 2\n<strong>Output:</strong> 91\n<strong>Explanation:</strong> The answer should be the total numbers in the range of 0 ≤ x < 100, excluding 11,22,33,44,55,66,77,88,99\n</pre>\n\n<p><strong>Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> n = 0\n<strong>Output:</strong> 1\n</pre>\n\n<p> </p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>0 <= n <= 8</code></li>\n</ul>\n",
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"translatedTitle": "统计各位数字都不同的数字个数",
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"translatedContent": "给你一个整数 <code>n</code> ,统计并返回各位数字都不同的数字 <code>x</code> 的个数,其中 <code>0 <= x < 10<sup>n</sup></code><sup> </sup>。\n<div class=\"original__bRMd\">\n<div>\n<p> </p>\n\n<p><strong>示例 1:</strong></p>\n\n<pre>\n<strong>输入:</strong>n = 2\n<strong>输出:</strong>91\n<strong>解释:</strong>答案应为除去 <code>11、22、33、44、55、66、77、88、99 </code>外,在 0 ≤ x < 100 范围内的所有数字。 \n</pre>\n\n<p><strong>示例 2:</strong></p>\n\n<pre>\n<strong>输入:</strong>n = 0\n<strong>输出:</strong>1\n</pre>\n</div>\n</div>\n\n<p> </p>\n\n<p><strong>提示:</strong></p>\n\n<ul>\n\t<li><code>0 <= n <= 8</code></li>\n</ul>\n",
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"A direct way is to use the backtracking approach.",
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"Backtracking should contains three states which are (the current number, number of steps to get that number and a bitmask which represent which number is marked as visited so far in the current number). Start with state (0,0,0) and count all valid number till we reach number of steps equals to 10<sup>n</sup>.",
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"This problem can also be solved using a dynamic programming approach and some knowledge of combinatorics.",
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"Let f(k) = count of numbers with unique digits with length equals k.",
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"f(1) = 10, ..., f(k) = 9 * 9 * 8 * ... (9 - k + 2) [The first factor is 9 because a number cannot start with 0]."
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