# 3Sum

15. 3Sum

LintCode-57.3Sum

Given an array S of n integers, are there elements a, b, c in S 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.

``````For example, given array S = [-1, 0, 1, 2, -1, -4],

A solution set is:
[
[-1, 0, 1],
[-1, -1, 2]
]
``````
``````class Solution {
func threeSum(_ nums: [Int]) -> [[Int]] {
let nums = nums.sorted { \$0 < \$1 }
var result: [[Int]] = []

if nums.count < 3 {
return []
}

for i in 0..<nums.count - 2 {
if i > 0 && nums[i] == nums[i - 1] {
continue
}

var low = i + 1, high = nums.count - 1
while low < high {
let sum = nums[i] + nums[low] + nums[high]

if sum == 0 {
result.append([nums[i], nums[low], nums[high]])

low += 1
high -= 1

while low < high && nums[low] == nums[low - 1] {
low += 1
}
while low < high && nums[high] == nums[high + 1] {
high -= 1
}
} else if sum < 0 {
low += 1
} else {
high -= 1
}
}
}

return result
}
}
``````
``````public class Solution {
public List<List<Integer>> threeSum(int[] nums) {
List<List<Integer>> result = new ArrayList<>();
if (nums == null || nums.length < 3) {
return result;
}

Arrays.sort(nums);

for (int i = 0; i < nums.length - 2; i++) {
if (i > 0 && nums[i] == nums[i - 1]) {
continue;
}

int low = i + 1, high = nums.length - 1;
while (low < high) {
int sum = nums[i] + nums[low] + nums[high];

if (sum == 0) {
List<Integer> list = new ArrayList<>();

low++;
high--;

while (low < high && nums[low] == nums[low - 1]) {
low++;
}

while (low < high && nums[high] == nums[high + 1]) {
high--;
}
} else if (sum < 0) {
low++;
} else {
high--;
}
}
}

return result;
}
}
``````

Hope this helps,
Michael