// LeetCode #15
// Difficulty: Medium
// For test
// Time:O(n^2)
// Space:O(n^2)
// Stl: SORT, ITERATOR, VECTOR


#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
int i, j, k, n;

//有點類似排序演算法的作法
//每一個輪使用一個index先固定一個數 而剩下的陣列內容 分別使用兩個不同的index左右逼近 
//過程中要注意避開重複的數字

int main() {

    int nums[] = { -1, 0, 1, 2, -1, -4 };
    //convenient to stores result by using two-dimensional vector
    vector<vector<int> > res;
    vector<int> temp;
    n = sizeof(nums) / sizeof(nums[0]);
    sort(nums, nums + n);

    for (int i = 0; i < 6; i++) {

        //jump over duplicate
        if (i > 0 && nums[i] == nums[i - 1])
            continue;
        //two indices on each for loop
        k = n - 1; j = i + 1;
        while (j < k) {
            if (nums[i] + nums[j] + nums[k] == 0)
            {
                temp.push_back(nums[i]);
                temp.push_back(nums[j]);
                temp.push_back(nums[k]);
                res.push_back(temp);
                while (j < k && nums[k] == nums[k - 1])k--;
                while (j < k && nums[j] == nums[j + 1])j++;
                k--; j++;
                temp.clear();
            }
            else if (nums[i] + nums[j] + nums[k] > 0)k--;
            else j++;
        }
    }

    for (int v = 0; v < res.size(); ++v) {

        for (vector<int>::iterator it = res[v].begin(); it != res[v].end(); ++it)
        {
            cout << *it << " ";
        }
        cout << endl;
    }

    system("PAUSE");
}



//SOLUTION

class Solution {
public:
    vector<vector<int>> threeSum(vector<int>& nums) {
        int i, j, k,n = nums.size();

        vector<vector<int> > res;
        vector<int> temp;

        sort(nums.begin(),nums.begin()+n);

        for (int i = 0; i < n; i++) {

            //jump over duplicate
            if (i > 0 && nums[i] == nums[i - 1])
                continue;
            //two indices on each for loop
            k = n - 1; j = i + 1;
            while (j < k) {
                if (nums[i] + nums[j] + nums[k] == 0)
                {
                    temp.push_back(nums[i]);
                    temp.push_back(nums[j]);
                    temp.push_back(nums[k]);
                    res.push_back(temp);
                    while (j < k && nums[k] == nums[k - 1])k--;
                    while (j < k && nums[j] == nums[j + 1])j++;
                    k--; j++;
                    temp.clear();
                }
                else if (nums[i] + nums[j] + nums[k] > 0)k--;
                else j++;
            }
        }
        return res;
    }
};