126 lines
3.3 KiB
Markdown
126 lines
3.3 KiB
Markdown
## 题目地址
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https://leetcode.com/problems/3sum/description/
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## 题目描述
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```
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Given an array nums of n integers, are there elements a, b, c in nums such that a + b + c = 0? Find all unique triplets in the array which gives the sum of zero.
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Note:
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The solution set must not contain duplicate triplets.
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Example:
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Given array nums = [-1, 0, 1, 2, -1, -4],
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A solution set is:
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[
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[-1, 0, 1],
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[-1, -1, 2]
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]
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```
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## 思路
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我们采用`分治`的思想. 想要找出三个数相加等于0，我们可以数组依次遍历，
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每一项a[i]我们都认为它是最终能够用组成0中的一个数字，那么我们的目标就是找到
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剩下的元素（除a[i]）`两个`相加等于-a[i].
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通过上面的思路，我们的问题转化为了`给定一个数组，找出其中两个相加等于给定值`，
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这个问题是比较简单的， 我们只需要对数组进行排序，然后双指针解决即可。 加上我们需要外层遍历依次数组，因此总的时间复杂度应该是O(N^2)。
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思路如图所示：
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> 在这里之所以要排序解决是因为， 我们算法的瓶颈在这里不在于排序，而在于O(N^2)，如果我们瓶颈是排序，就可以考虑别的方式了
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> 如果找某一个特定元素，一个指针就够了。如果是找两个元素满足一定关系（比如求和等于特定值），需要双指针，
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当然前提是数组有序。
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## 关键点解析
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- 排序之后，用双指针
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- 分治
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## 代码
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```js
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/*
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* @lc app=leetcode id=15 lang=javascript
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*
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* [15] 3Sum
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*
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* https://leetcode.com/problems/3sum/description/
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*
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* algorithms
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* Medium (23.51%)
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* Total Accepted: 531.5K
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* Total Submissions: 2.2M
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* Testcase Example: '[-1,0,1,2,-1,-4]'
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*
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* Given an array nums of n integers, are there elements a, b, c in nums such
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* that a + b + c = 0? Find all unique triplets in the array which gives the
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* sum of zero.
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*
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* Note:
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*
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* The solution set must not contain duplicate triplets.
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*
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* Example:
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*
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*
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* Given array nums = [-1, 0, 1, 2, -1, -4],
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*
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* A solution set is:
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* [
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* [-1, 0, 1],
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* [-1, -1, 2]
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* ]
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*
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*
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*/
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/**
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* @param {number[]} nums
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* @return {number[][]}
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*/
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var threeSum = function(nums) {
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if (nums.length < 3) return [];
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const list = [];
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nums.sort((a, b) => a - b);
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for (let i = 0; i < nums.length; i++) {
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//nums is sorted,so it's impossible to have a sum = 0
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if (nums[i] > 0) break;
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// skip duplicated result without set
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if (i > 0 && nums[i] === nums[i - 1]) continue;
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let left = i + 1;
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let right = nums.length - 1;
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// for each index i
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// we want to find the triplet [i, left, right] which sum to 0
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while (left < right) {
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// since left < right, and left > i, no need to compare i === left and i === right.
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if (nums[left] + nums[right] + nums[i] === 0) {
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list.push([nums[left], nums[right], nums[i]]);
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// skip duplicated result without set
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while(nums[left] === nums[left + 1]) {
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left++;
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}
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left++;
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// skip duplicated result without set
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while(nums[right] === nums[right - 1]) {
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right--;
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}
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right--;
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continue;
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} else if (nums[left] + nums[right] + nums[i] > 0) {
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right--;
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} else {
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left++;
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}
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}
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}
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return list;
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};
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```
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