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98 lines
3.2 KiB
Markdown
98 lines
3.2 KiB
Markdown
# [18. 4Sum](https://leetcode.com/problems/4sum/)
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## 题目
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Given an array nums of n integers and an integer target, are there elements a, b, c, and d in nums such that a + b + c + d = target? Find all unique quadruplets in the array which gives the sum of target.
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**Note**:
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The solution set must not contain duplicate quadruplets.
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**Example**:
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```
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Given array nums = [1, 0, -1, 0, -2, 2], and target = 0.
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A solution set is:
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[
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[-1, 0, 0, 1],
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[-2, -1, 1, 2],
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[-2, 0, 0, 2]
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]
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```
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## 题目大意
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给定一个数组，要求在这个数组中找出 4 个数之和为 0 的所有组合。
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## 解题思路
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用 map 提前计算好任意 3 个数字之和，保存起来，可以将时间复杂度降到 O(n^3)。这一题比较麻烦的一点在于，最后输出解的时候，要求输出不重复的解。数组中同一个数字可能出现多次，同一个数字也可能使用多次，但是最后输出解的时候，不能重复。例如 [-1，1，2, -2] 和 [2, -1, -2, 1]、[-2, 2, -1, 1] 这 3 个解是重复的，即使 -1, -2 可能出现 100 次，每次使用的 -1, -2 的数组下标都是不同的。
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这一题是第 15 题的升级版，思路都是完全一致的。这里就需要去重和排序了。map 记录每个数字出现的次数，然后对 map 的 key 数组进行排序，最后在这个排序以后的数组里面扫，找到另外 3 个数字能和自己组成 0 的组合。
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第 15 题和第 18 题的解法一致。
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## 代码
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```go
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package leetcode
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import "sort"
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func fourSum(nums []int, target int) [][]int {
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res := [][]int{}
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counter := map[int]int{}
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for _, value := range nums {
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counter[value]++
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}
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uniqNums := []int{}
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for key := range counter {
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uniqNums = append(uniqNums, key)
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}
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sort.Ints(uniqNums)
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for i := 0; i < len(uniqNums); i++ {
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if (uniqNums[i]*4 == target) && counter[uniqNums[i]] >= 4 {
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res = append(res, []int{uniqNums[i], uniqNums[i], uniqNums[i], uniqNums[i]})
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}
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for j := i + 1; j < len(uniqNums); j++ {
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if (uniqNums[i]*3+uniqNums[j] == target) && counter[uniqNums[i]] > 2 {
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res = append(res, []int{uniqNums[i], uniqNums[i], uniqNums[i], uniqNums[j]})
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}
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if (uniqNums[j]*3+uniqNums[i] == target) && counter[uniqNums[j]] > 2 {
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res = append(res, []int{uniqNums[i], uniqNums[j], uniqNums[j], uniqNums[j]})
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}
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if (uniqNums[j]*2+uniqNums[i]*2 == target) && counter[uniqNums[j]] > 1 && counter[uniqNums[i]] > 1 {
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res = append(res, []int{uniqNums[i], uniqNums[i], uniqNums[j], uniqNums[j]})
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}
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for k := j + 1; k < len(uniqNums); k++ {
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if (uniqNums[i]*2+uniqNums[j]+uniqNums[k] == target) && counter[uniqNums[i]] > 1 {
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res = append(res, []int{uniqNums[i], uniqNums[i], uniqNums[j], uniqNums[k]})
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}
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if (uniqNums[j]*2+uniqNums[i]+uniqNums[k] == target) && counter[uniqNums[j]] > 1 {
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res = append(res, []int{uniqNums[i], uniqNums[j], uniqNums[j], uniqNums[k]})
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}
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if (uniqNums[k]*2+uniqNums[i]+uniqNums[j] == target) && counter[uniqNums[k]] > 1 {
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res = append(res, []int{uniqNums[i], uniqNums[j], uniqNums[k], uniqNums[k]})
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}
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c := target - uniqNums[i] - uniqNums[j] - uniqNums[k]
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if c > uniqNums[k] && counter[c] > 0 {
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res = append(res, []int{uniqNums[i], uniqNums[j], uniqNums[k], c})
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}
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}
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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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