104 lines
2.4 KiB
Markdown
104 lines
2.4 KiB
Markdown
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## 题目地址
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https://leetcode.com/problems/4sum-ii/description/
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## 题目描述
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```
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Given four lists A, B, C, D of integer values, compute how many tuples (i, j, k, l) there are such that A[i] + B[j] + C[k] + D[l] is zero.
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To make problem a bit easier, all A, B, C, D have same length of N where 0 ≤ N ≤ 500. All integers are in the range of -228 to 228 - 1 and the result is guaranteed to be at most 231 - 1.
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Example:
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Input:
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A = [ 1, 2]
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B = [-2,-1]
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C = [-1, 2]
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D = [ 0, 2]
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Output:
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2
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Explanation:
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The two tuples are:
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1. (0, 0, 0, 1) -> A[0] + B[0] + C[0] + D[1] = 1 + (-2) + (-1) + 2 = 0
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2. (1, 1, 0, 0) -> A[1] + B[1] + C[0] + D[0] = 2 + (-1) + (-1) + 0 = 0
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```
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## 思路
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如果按照常规思路去完成查找需要四层遍历，时间复杂是O(n^4), 显然是行不通的。
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因此我们有必要想一种更加高效的算法。
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我一个思路就是我们将四个数组分成两组，两两结合。
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然后我们分别计算`两两结合能够算出的和有哪些，以及其对应的个数`。
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如图：
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这个时候我们得到了两个`hashTable`， 我们只需要进行简单的数学运算就可以得到结果。
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## 关键点解析
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- 空间换时间
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- 两两分组，求出两两结合能够得出的可能数，然后合并即可。
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## 代码
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语言支持： `JavaScript`，`Python3`
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`JavaScript`:
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```js
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/*
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* @lc app=leetcode id=454 lang=javascript
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*
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* [454] 4Sum II
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*
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* https://leetcode.com/problems/4sum-ii/description/
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/**
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* @param {number[]} A
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* @param {number[]} B
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* @param {number[]} C
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* @param {number[]} D
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* @return {number}
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*/
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var fourSumCount = function(A, B, C, D) {
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const sumMapper = {};
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let res = 0;
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for (let i = 0; i < A.length; i++) {
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for (let j = 0; j < B.length; j++) {
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sumMapper[A[i] + B[j]] = (sumMapper[A[i] + B[j]] || 0) + 1;
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}
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}
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for (let i = 0; i < C.length; i++) {
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for (let j = 0; j < D.length; j++) {
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res += sumMapper[- (C[i] + D[j])] || 0;
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}
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}
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return res;
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};
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```
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`Python3`:
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```python
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class Solution:
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def fourSumCount(self, A: List[int], B: List[int], C: List[int], D: List[int]) -> int:
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mapper = {}
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res = 0
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for i in A:
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for j in B:
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mapper[i + j] = mapper.get(i + j, 0) + 1
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for i in C:
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for j in D:
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res += mapper.get(-1 * (i + j), 0)
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return res
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```
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