【Leetcode】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:

• 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)


2 sum 变化而来的题目，首先第一个元素逐个循环，后面的部分采用2sum的思路双指针逼近即可。

1st （12 tries）

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> > re;
        if(num.size() < 3)
            return re;
        sort(num.begin(),num.end());
        for(int i = 0;i < num.size() - 2;i++)
        {
            vector<int> tmp;
            tmp.push_back(num[i]);
            int num1 = num[i];
            /*important*/
            if(i > 0 && num[i] == num[i - 1])
            {
                continue;
            }
            int left = i + 1;
            int right = num.size() - 1;
            int preleft = 1024,preright = 1024;
            while(right > left)
            {
                if(num1 + num[right] + num[left] == 0)
                {
                    /*important*/
                    if(preleft == num[left] && preright == num[right])
                    {
                        left ++;
                        right --;
                    }
                    else
                    {
                        tmp.push_back(num[left]);
                        tmp.push_back(num[right]);
                        preleft = num[left];
                        preright = num[right];
                        re.push_back(tmp);
                        tmp.pop_back();
                        tmp.pop_back();
                        left ++;
                        right --;
                    }
                }
                if(num1 + num[right] + num[left] < 0)
                {
                    left ++;
                }
                if(num1 + num[right] + num[left] > 0)
                {
                    right --;
                }
            }
        }
        return re;
    }
};

　　

2nd （5 tries）

class Solution {
public:
    vector<vector<int> > ans;
    vector<int> res;
    vector<vector<int> > threeSum(vector<int> &num) {
        sort(num.begin(),num.end());
        //scan one with a 2-sum problem
        int n = num.size();
        for(int first = 0;first < n-2;first++) {
            //remove duplicated
            if(first > 0 && num[first] == num[first-1])
                continue;
            //2-sum
            int start = first+1;
            int end = n - 1;
            while(start < end) {
                //remove dup
                if( start > first+1 && num[start] == num[start-1] && end < n - 1 && num[end] == num[end + 1]) {
                    start++;
                    end--;
                    continue;
                }
                if(num[start] + num[end] + num[first] == 0) {
                    res.push_back(num[first]);
                    res.push_back(num[start]);
                    res.push_back(num[end]);
                    ans.push_back(res);
                    res.clear();
                    start++;
                    end--;
                }
                else if(num[start] + num[end] + num[first] > 0) {
                    end--;
                } 
                else {
                    start++;
                }
            }
        }
        return ans;
    }
};