Given an array nums of n integers, are there elements a, b, c in numssuch 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]
]
Solution:
Loop through i and do a standard 2 sum. Use while loops to skip duplicate values
class Solution {
public:
vector<vector<int>> threeSum(vector<int>& nums) {
vector<vector<int> > ans;
sort(nums.begin(),nums.end());
for(int i=0;i<nums.size();i++){
int a = i+1;
int b = nums.size()-1;
while(a<b){
int target = -nums[i];
int x = nums[a], y=nums[b];
if(x+y>target) b--;
else if(x+y<target)a++;
else{
ans.push_back(vector<int>{nums[i],x,y});
while(a+1<nums.size()&&nums[a+1]==x) a++;
while(b-1>i&&nums[b-1]==y)b--;
a++;b--;
}
}
while(i+1<nums.size()&&nums[i+1]==nums[i])i++;
}
return ans;
}
};