[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)
分析：如果使用三层循环的话肯定会Time Limited Exceeded的！

class Solution {
public:
    vector<vector<int> > threeSum(vector<int> &num) {
        vector<vector<int> > result;
        int n = (int)num.size();
        if (n > 2) {
            sort(num.begin(), num.end());

            for (int i = 1; i < n - 1; ++i) {//序号i为中间的数，序号l为较小的数，序号r为较大的数，每次循环时先固定一个中间的数i
                if (i > 2 && num[i] == num[i - 2]) 
                    continue;
                int l = 0, r = n - 1;
                while (l < i && r > i) {
                    if (l > 1 && num[l] == num[l - 1]) {
                        ++l; 
                        continue;
                    }
                    if (r < n - 1 && num[r] == num[r + 1]) {
                        --r; 
                        continue;
                    }
                    int sum = num[l] + num[i] + num[r];
                    if (sum == 0) {
                        vector<int> item;
                        item.push_back( num[l]);
                        item.push_back( num[i]);
                        item.push_back( num[r]);
                        bool flag = false;
                        for (int j = 0; j < result.size(); ++j) {
                            if (result[j][0] == num[l] && result[j][1] == num[i]) {
                                flag = true; 
                                break;
                            }
                        }
                        if (!flag) 
                            result.push_back(item);
                        ++l; 
                        --r;
                    } 
                    else if (sum < 0)
                        ++l; 
                    else 
                        --r;
                }//end while
            }//end for
        }//end if
        return result;
    }

};