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程序员小墨 2023-12-09 19:57:46 +08:00
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1
get.bat
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git checkout master
git pull
git count-objects -vH
python leetcode.py
python leetcode-cn.py

4280
leetcode-cn/origin-data.json
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leetcode-cn/originData/3sum.json
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"translatedContent": "<p>给你一个整数数组 <code>nums</code> ，判断是否存在三元组 <code>[nums[i], nums[j], nums[k]]</code> 满足 <code>i != j</code>、<code>i != k</code> 且 <code>j != k</code> ，同时还满足 <code>nums[i] + nums[j] + nums[k] == 0</code> 。请</p>\n\n<p>你返回所有和为 <code>0</code> 且不重复的三元组。</p>\n\n<p><strong>注意：</strong>答案中不可以包含重复的三元组。</p>\n\n<p>&nbsp;</p>\n\n<p>&nbsp;</p>\n\n<p><strong>示例 1：</strong></p>\n\n<pre>\n<strong>输入：</strong>nums = [-1,0,1,2,-1,-4]\n<strong>输出：</strong>[[-1,-1,2],[-1,0,1]]\n<strong>解释：</strong>\nnums[0] + nums[1] + nums[2] = (-1) + 0 + 1 = 0 。\nnums[1] + nums[2] + nums[4] = 0 + 1 + (-1) = 0 。\nnums[0] + nums[3] + nums[4] = (-1) + 2 + (-1) = 0 。\n不同的三元组是 [-1,0,1] 和 [-1,-1,2] 。\n注意，输出的顺序和三元组的顺序并不重要。\n</pre>\n\n<p><strong>示例 2：</strong></p>\n\n<pre>\n<strong>输入：</strong>nums = [0,1,1]\n<strong>输出：</strong>[]\n<strong>解释：</strong>唯一可能的三元组和不为 0 。\n</pre>\n\n<p><strong>示例 3：</strong></p>\n\n<pre>\n<strong>输入：</strong>nums = [0,0,0]\n<strong>输出：</strong>[[0,0,0]]\n<strong>解释：</strong>唯一可能的三元组和为 0 。\n</pre>\n\n<p>&nbsp;</p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li><code>3 &lt;= nums.length &lt;= 3000</code></li>\n\t<li><code>-10<sup>5</sup> &lt;= nums[i] &lt;= 10<sup>5</sup></code></li>\n</ul>\n",
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"So, we essentially need to find three numbers x, y, and z such that they add up to the given value. If we fix one of the numbers say x, we are left with the two-sum problem at hand!",
"For the two-sum problem, if we fix one of the numbers, say x, we have to scan the entire array to find the next number y, which is value - x where value is the input parameter. Can we change our array somehow so that this search becomes faster?",

2
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leetcode-cn/originData/4sum.json
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"translatedContent": "<p>给你一个由 <code>n</code> 个整数组成的数组&nbsp;<code>nums</code> ，和一个目标值 <code>target</code> 。请你找出并返回满足下述全部条件且<strong>不重复</strong>的四元组&nbsp;<code>[nums[a], nums[b], nums[c], nums[d]]</code>&nbsp;（若两个四元组元素一一对应，则认为两个四元组重复）：</p>\n\n<ul>\n\t<li><code>0 &lt;= a, b, c, d&nbsp;&lt; n</code></li>\n\t<li><code>a</code>、<code>b</code>、<code>c</code> 和 <code>d</code> <strong>互不相同</strong></li>\n\t<li><code>nums[a] + nums[b] + nums[c] + nums[d] == target</code></li>\n</ul>\n\n<p>你可以按 <strong>任意顺序</strong> 返回答案 。</p>\n\n<p>&nbsp;</p>\n\n<p><strong>示例 1：</strong></p>\n\n<pre>\n<strong>输入：</strong>nums = [1,0,-1,0,-2,2], target = 0\n<strong>输出：</strong>[[-2,-1,1,2],[-2,0,0,2],[-1,0,0,1]]\n</pre>\n\n<p><strong>示例 2：</strong></p>\n\n<pre>\n<strong>输入：</strong>nums = [2,2,2,2,2], target = 8\n<strong>输出：</strong>[[2,2,2,2]]\n</pre>\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;= 200</code></li>\n\t<li><code>-10<sup>9</sup> &lt;= nums[i] &lt;= 10<sup>9</sup></code></li>\n\t<li><code>-10<sup>9</sup> &lt;= target &lt;= 10<sup>9</sup></code></li>\n</ul>\n",
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"translatedContent": "<p>以数组 <code>intervals</code> 表示若干个区间的集合，其中单个区间为 <code>intervals[i] = [start<sub>i</sub>, end<sub>i</sub>]</code> 。请你合并所有重叠的区间，并返回一个不重叠的区间数组，该数组需恰好覆盖输入中的所有区间。</p>\n\n<p>&nbsp;</p>\n\n<p><strong>示例 1：</strong></p>\n\n<pre>\n<strong>输入：</strong>intervals = [[1,3],[2,6],[8,10],[15,18]]\n<strong>输出：</strong>[[1,6],[8,10],[15,18]]\n<strong>解释：</strong>区间 [1,3] 和 [2,6] 重叠, 将它们合并为 [1,6].\n</pre>\n\n<p><strong>示例&nbsp;2：</strong></p>\n\n<pre>\n<strong>输入：</strong>intervals = [[1,4],[4,5]]\n<strong>输出：</strong>[[1,5]]\n<strong>解释：</strong>区间 [1,4] 和 [4,5] 可被视为重叠区间。</pre>\n\n<p>&nbsp;</p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= intervals.length &lt;= 10<sup>4</sup></code></li>\n\t<li><code>intervals[i].length == 2</code></li>\n\t<li><code>0 &lt;= start<sub>i</sub> &lt;= end<sub>i</sub> &lt;= 10<sup>4</sup></code></li>\n</ul>\n\n<p>&nbsp;</p>\n\n<p><meta charset=\"UTF-8\" />注意：本题与主站 56&nbsp;题相同：&nbsp;<a href=\"https://leetcode-cn.com/problems/merge-intervals/\">https://leetcode-cn.com/problems/merge-intervals/</a></p>\n",
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