15 3Sum – Medium
Problem:
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:
Elements in a triplet (a,b,c) must be in non-descending order. (ie, a ≤ b ≤ c) 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)
Thoughts:
Key idea here is to sort the array first. Then start from left to pick the first element to be a, then find if there are two other elements b, c make a + b + c = 0. Another thing to keep in mind is that how to avoid duplicate answers.
Solutions:
public class Solution {
public List<List<Integer>> threeSum(int[] nums) {
List<List<Integer>> result = new LinkedList<List<Integer>>();
Arrays.sort(nums);
for (int i = 0; i < nums.length; i++) {
if (i == 0 || nums[i] != nums[i-1]) {
int start = i + 1, end = nums.length - 1;
while (start < end) {
int sum = nums[i] + nums[start] + nums[end];
if (sum == 0){
List<Integer> tmp = new LinkedList<Integer>();
tmp.add(nums[i]);
tmp.add(nums[start]);
tmp.add(nums[end]);
result.add(tmp);
int startVal = nums[start];
int endVal = nums[end];
while (start < end && startVal == nums[start]) {
start ++;
}
while (end > start && endVal == nums[end]) {
end --;
}
}
else if (sum < 0) {
start ++;
}
else {
end --;
}
}
}
}
return result;
}
}