0018.4 Sum

# 18. 4Sum#

## 题目 #

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.

Note:

The solution set must not contain duplicate quadruplets.

Example:

``````
Given array nums = [1, 0, -1, 0, -2, 2], and target = 0.

A solution set is:
[
[-1,  0, 0, 1],
[-2, -1, 1, 2],
[-2,  0, 0, 2]
]

``````

## 代码 #

``````
package leetcode

import "sort"

func fourSum(nums []int, target int) [][]int {
res := [][]int{}
counter := map[int]int{}
for _, value := range nums {
counter[value]++
}

uniqNums := []int{}
for key := range counter {
uniqNums = append(uniqNums, key)
}
sort.Ints(uniqNums)

for i := 0; i < len(uniqNums); i++ {
if (uniqNums[i]*4 == target) && counter[uniqNums[i]] >= 4 {
res = append(res, []int{uniqNums[i], uniqNums[i], uniqNums[i], uniqNums[i]})
}
for j := i + 1; j < len(uniqNums); j++ {
if (uniqNums[i]*3+uniqNums[j] == target) && counter[uniqNums[i]] > 2 {
res = append(res, []int{uniqNums[i], uniqNums[i], uniqNums[i], uniqNums[j]})
}
if (uniqNums[j]*3+uniqNums[i] == target) && counter[uniqNums[j]] > 2 {
res = append(res, []int{uniqNums[i], uniqNums[j], uniqNums[j], uniqNums[j]})
}
if (uniqNums[j]*2+uniqNums[i]*2 == target) && counter[uniqNums[j]] > 1 && counter[uniqNums[i]] > 1 {
res = append(res, []int{uniqNums[i], uniqNums[i], uniqNums[j], uniqNums[j]})
}
for k := j + 1; k < len(uniqNums); k++ {
if (uniqNums[i]*2+uniqNums[j]+uniqNums[k] == target) && counter[uniqNums[i]] > 1 {
res = append(res, []int{uniqNums[i], uniqNums[i], uniqNums[j], uniqNums[k]})
}
if (uniqNums[j]*2+uniqNums[i]+uniqNums[k] == target) && counter[uniqNums[j]] > 1 {
res = append(res, []int{uniqNums[i], uniqNums[j], uniqNums[j], uniqNums[k]})
}
if (uniqNums[k]*2+uniqNums[i]+uniqNums[j] == target) && counter[uniqNums[k]] > 1 {
res = append(res, []int{uniqNums[i], uniqNums[j], uniqNums[k], uniqNums[k]})
}
c := target - uniqNums[i] - uniqNums[j] - uniqNums[k]
if c > uniqNums[k] && counter[c] > 0 {
res = append(res, []int{uniqNums[i], uniqNums[j], uniqNums[k], c})
}
}
}
}
return res
}

``````

Sep 6, 2020