leetcode-problemset/leetcode-cn/originData/3sum.json

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            "title": "3Sum",
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            "content": "<p>Given an integer array nums, return all the triplets <code>[nums[i], nums[j], nums[k]]</code> such that <code>i != j</code>, <code>i != k</code>, and <code>j != k</code>, and <code>nums[i] + nums[j] + nums[k] == 0</code>.</p>\n\n<p>Notice that the solution set must not contain duplicate triplets.</p>\n\n<p>&nbsp;</p>\n<p><strong class=\"example\">Example 1:</strong></p>\n\n<pre>\n<strong>Input:</strong> nums = [-1,0,1,2,-1,-4]\n<strong>Output:</strong> [[-1,-1,2],[-1,0,1]]\n<strong>Explanation:</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.\nThe distinct triplets are [-1,0,1] and [-1,-1,2].\nNotice that the order of the output and the order of the triplets does not matter.\n</pre>\n\n<p><strong class=\"example\">Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> nums = [0,1,1]\n<strong>Output:</strong> []\n<strong>Explanation:</strong> The only possible triplet does not sum up to 0.\n</pre>\n\n<p><strong class=\"example\">Example 3:</strong></p>\n\n<pre>\n<strong>Input:</strong> nums = [0,0,0]\n<strong>Output:</strong> [[0,0,0]]\n<strong>Explanation:</strong> The only possible triplet sums up to 0.\n</pre>\n\n<p>&nbsp;</p>\n<p><strong>Constraints:</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",
            "translatedTitle": "三数之和",
            "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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            "hints": [
                "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?",
                "The second train of thought for two-sum is, without changing the array, can we use additional space somehow? Like maybe a hash map to speed up the search?"
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