LeetCode 15: 3Sum

Difficulty: 3

Frequency: 5

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)

Solution:

class Solution {public:    vector<vector<int> > threeSum(vector<int> &num) {        // Start typing your C/C++ solution below        // DO NOT write int main() function        vector<vector<int> > answer;        if (num.size()<3)            return answer;                vector<int> triplet;        sort(num.begin(), num.end());        int i_start, i_end;        int sum;        for (int i = 0; i<num.size()-2; ++i)        {            if (i>0&&num[i]==num[i-1])                continue;                            i_start = i+1;            i_end = num.size()-1;                        while (i_start<i_end)            {                if (num[i_start]==num[i_start+1]&&num[i] + num[i_start] + num[i_start+1]==0)                {                    triplet.clear();                    triplet.push_back(num[i]);                    triplet.push_back(num[i_start]);                    triplet.push_back(num[i_start]);                    answer.push_back(triplet);                }                while (i_start<i_end&&num[i_start]==num[i_start+1])                    ++i_start;                                    if (i_start==i_end)                    break;                                    if (num[i_end]==num[i_end-1]&&num[i] + num[i_end-1] + num[i_end]==0)                {                    triplet.clear();                    triplet.push_back(num[i]);                    triplet.push_back(num[i_end]);                    triplet.push_back(num[i_end]);                    answer.push_back(triplet);                }                while (i_end>i_start&&num[i_end]==num[i_end-1])                    --i_end;                                    sum = num[i] + num[i_start] + num[i_end];                if (sum==0)                {                    triplet.clear();                    triplet.push_back(num[i]);                    triplet.push_back(num[i_start]);                    triplet.push_back(num[i_end]);                    answer.push_back(triplet);                    ++i_start;                }                else if (sum>0)                    --i_end;                else                    ++i_start;            }        }        return answer;    }};

Notes:

If you know how to do 2Sum, then it is n rounds of 2Sum.

I think the biggest difficulty is how to remove duplicated answers.

I will consult others' code for other solution.