0015.3 Sum

# 15. 3Sum#

## 题目 #

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.

Note:

The solution set must not contain duplicate triplets.

Example:

``````
Given array nums = [-1, 0, 1, 2, -1, -4],

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

``````

## 代码 #

``````
package leetcode

import (
"sort"
)

// 解法一 最优解，双指针 + 排序
func threeSum(nums []int) [][]int {
sort.Ints(nums)
result, start, end, index, addNum, length := make([][]int, 0), 0, 0, 0, 0, len(nums)
for index = 1; index < length-1; index++ {
start, end = 0, length-1
if index > 1 && nums[index] == nums[index-1] {
start = index - 1
}
for start < index && end > index {
if start > 0 && nums[start] == nums[start-1] {
start++
continue
}
if end < length-1 && nums[end] == nums[end+1] {
end--
continue
}
addNum = nums[start] + nums[end] + nums[index]
result = append(result, []int{nums[start], nums[index], nums[end]})
start++
end--
} else if addNum > 0 {
end--
} else {
start++
}
}
}
return result
}

// 解法二
func threeSum1(nums []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]*3 == 0) && counter[uniqNums[i]] >= 3 {
res = append(res, []int{uniqNums[i], uniqNums[i], uniqNums[i]})
}
for j := i + 1; j < len(uniqNums); j++ {
if (uniqNums[i]*2+uniqNums[j] == 0) && counter[uniqNums[i]] > 1 {
res = append(res, []int{uniqNums[i], uniqNums[i], uniqNums[j]})
}
if (uniqNums[j]*2+uniqNums[i] == 0) && counter[uniqNums[j]] > 1 {
res = append(res, []int{uniqNums[i], uniqNums[j], uniqNums[j]})
}
c := 0 - uniqNums[i] - uniqNums[j]
if c > uniqNums[j] && counter[c] > 0 {
res = append(res, []int{uniqNums[i], uniqNums[j], c})
}
}
}
return res
}

``````

Nov 25, 2022