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"translatedContent": "<p>给你一个包含 <code>n</code> 个整数的数组 <code>nums</code>,判断 <code>nums</code> 中是否存在三个元素 <em>a,b,c ,</em>使得 <em>a + b + c = </em>0 ?请你找出所有和为 <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> </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</pre>\n\n<p><strong>示例 2:</strong></p>\n\n<pre>\n<strong>输入:</strong>nums = []\n<strong>输出:</strong>[]\n</pre>\n\n<p><strong>示例 3:</strong></p>\n\n<pre>\n<strong>输入:</strong>nums = [0]\n<strong>输出:</strong>[]\n</pre>\n\n<p> </p>\n\n<p><strong>提示:</strong></p>\n\n<ul>\n\t<li><code>0 <= nums.length <= 3000</code></li>\n\t<li><code>-10<sup>5</sup> <= nums[i] <= 10<sup>5</sup></code></li>\n</ul>\n",
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"hints": [
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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!",
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"For the two-sum problem, if we fix one of the numbers, say <pre>x</pre>, we have to scan the entire array to find the next number<pre>y</pre> which is <pre>value - x</pre> where value is the input parameter. Can we change our array somehow so that this search becomes faster?",
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