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)

Solution:

Code:

<span style="font-size:14px;">class Solution {
public:
    vector<vector<int> > threeSum(vector<int> &num) {
        const int length = num.size();
        vector<vector<int> > results;
        sort(num.begin(), num.end());
        for (int i = 0; i < length-2; ++i) {
            int m = i+1, n = length-1;
            while (m < n) {
                if (num[i]+num[m]+num[n] == 0) {
                    vector<int> result;
                    result.push_back(num[i]);
                    result.push_back(num[m]);
                    result.push_back(num[n]);
                    results.push_back(result);
                    while (m+1 < n && num[m+1] == num[m]) ++m;
                    ++m;
                    while (n-1 > m && num[n-1] == num[n]) --n;
                    --n;
                } else if (num[i]+num[m]+num[n] < 0) {
                    while (m+1 < n && num[m+1] == num[m]) ++m;
                    ++m;
                } else if (num[i]+num[m]+num[n] > 0) {
                    while (n-1 > m && num[n-1] == num[n]) --n;
                    --n;
                }
            }
            while (i+1 < length-2 && num[i+1] == num[i]) ++i;
        }
        return results;
    }
};</span>