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]
]
题目大意 #
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。
解题思路 #
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。
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。
第 15 题和第 18 题的解法一致。
代码 #
package leetcode
import "sort"
// 解法一 双指针
func fourSum(nums []int, target int) (quadruplets [][]int) {
sort.Ints(nums)
n := len(nums)
for i := 0; i < n-3 && nums[i]+nums[i+1]+nums[i+2]+nums[i+3] <= target; i++ {
if i > 0 && nums[i] == nums[i-1] || nums[i]+nums[n-3]+nums[n-2]+nums[n-1] < target {
continue
}
for j := i + 1; j < n-2 && nums[i]+nums[j]+nums[j+1]+nums[j+2] <= target; j++ {
if j > i+1 && nums[j] == nums[j-1] || nums[i]+nums[j]+nums[n-2]+nums[n-1] < target {
continue
}
for left, right := j+1, n-1; left < right; {
if sum := nums[i] + nums[j] + nums[left] + nums[right]; sum == target {
quadruplets = append(quadruplets, []int{nums[i], nums[j], nums[left], nums[right]})
for left++; left < right && nums[left] == nums[left-1]; left++ {
}
for right--; left < right && nums[right] == nums[right+1]; right-- {
}
} else if sum < target {
left++
} else {
right--
}
}
}
}
return
}
// 解法二 kSum
func fourSum1(nums []int, target int) [][]int {
res, cur := make([][]int, 0), make([]int, 0)
sort.Ints(nums)
kSum(nums, 0, len(nums)-1, target, 4, cur, &res)
return res
}
func kSum(nums []int, left, right int, target int, k int, cur []int, res *[][]int) {
if right-left+1 < k || k < 2 || target < nums[left]*k || target > nums[right]*k {
return
}
if k == 2 {
// 2 sum
twoSum(nums, left, right, target, cur, res)
} else {
for i := left; i < len(nums); i++ {
if i == left || (i > left && nums[i-1] != nums[i]) {
next := make([]int, len(cur))
copy(next, cur)
next = append(next, nums[i])
kSum(nums, i+1, len(nums)-1, target-nums[i], k-1, next, res)
}
}
}
}
func twoSum(nums []int, left, right int, target int, cur []int, res *[][]int) {
for left < right {
sum := nums[left] + nums[right]
if sum == target {
cur = append(cur, nums[left], nums[right])
temp := make([]int, len(cur))
copy(temp, cur)
*res = append(*res, temp)
// reset cur to previous state
cur = cur[:len(cur)-2]
left++
right--
for left < right && nums[left] == nums[left-1] {
left++
}
for left < right && nums[right] == nums[right+1] {
right--
}
} else if sum < target {
left++
} else {
right--
}
}
}
// 解法三
func fourSum2(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
}