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)

Have you been asked this question in an interview?

我的解法是2Sum升级版，先按住一个数，然后做剩下的其他数的2Sum。注意之前按住的数之后可以不用考虑。复杂度为n*n。

因为要求所有的组合，所以num[i]+num[j]>sum时也要继续往下走，而且要略过所有相同的数。

vector<vector<int> > twoSum(vector<int> &num, int sum){
    vector<vector<int>> result;
    int i=0;
    int j=num.size()-1;
    while(i<j){
        if(num[i]+num[j]<sum){
            i++;
        }
        else if(num[i]+num[j]>sum){
            j--;
        }
        else{
            vector<int> tmp;
            tmp.push_back(num[i]);
            tmp.push_back(num[j]);
            result.push_back(tmp);
            int curi=num[i];
            int curj=num[j];
            while(num[++i]==curi);
            while(num[--j]==curj);
        }
    }
    return result;
}
vector<vector<int> > threeSum(vector<int> &num){
    vector<vector<int>> result;
    if(num.size()<3)
        return result;
    sort(num.begin(),num.end());
    for(int i=0;i<num.size();){
        vector<int> twosuminput(num.begin()+i+1,num.end());
        vector<vector<int>> twosumresult=twoSum(twosuminput,-num[i]);
        for(int j=0;j<twosumresult.size();j++){
            vector<int> tmp;
            tmp.push_back(num[i]);
            tmp.insert(tmp.begin()+1,twosumresult[j].begin(),twosumresult[j].end());
            result.push_back(tmp);
        }
        int cur=num[i];
        while(num[++i]==cur);
    }
    return result;
}