[LintCode 57] three Sum

LintCode 57. 三数之和

给出一个有n个整数的数组S，在S中找到三个整数a, b, c，找到所有使得a + b + c = 0的三元组。

样例
例1:

输入:[2,7,11,15]
输出:[]
例2:

输入:[-1,0,1,2,-1,-4]
输出:[[-1, 0, 1],[-1, -1, 2]]
注意事项
在三元组(a, b, c)，要求a <= b <= c。

结果不能包含重复的三元组。

暴力解法，时间复杂度O(N^3)，空间复杂度O(1)：

class Solution {
public:
    /**
     * @param numbers: Give an array numbers of n integer
     * @return: Find all unique triplets in the array which gives the sum of zero.
     */
    vector<vector<int>> threeSum(vector<int> &numbers) {
        // write your code here
        vector<vector<int>> res;
        std::sort(numbers.begin(), numbers.end());
        size_t n = numbers.size();
        int i=0, j,k;
        while (i < n-2) {
            j = i+1;
            while (j < n-1) {
                k = j+1;
                while (k < n) {
                    if (numbers[i] + numbers[j] + numbers[k] == 0) {
                        res.push_back({numbers[i], numbers[j], numbers[k]});
                        break;
                    }
                    k++;
                    while(k<n && numbers[k]==numbers[k-1]) k++;
                }
                j++;
                while (j<n-1 && numbers[j]==numbers[j-1]) j++;
            }
            i++;
            while (i<n-2 && numbers[i] == numbers[i-1]) i++;
        }
        return res;
    }
};

双指针解法，时间复杂度O(N^2)，空间复杂度O(1)：

class Solution {
public:
    /**
     * @param numbers: Give an array numbers of n integer
     * @return: Find all unique triplets in the array which gives the sum of zero.
     */
    vector<vector<int>> threeSum(vector<int> &numbers) {
        // write your code here
        vector<vector<int>> res;
        std::sort(numbers.begin(), numbers.end());
        size_t n = numbers.size();
        int i=0, j,k;
        while (i < n-2) {
            j = i + 1;
            k = n - 1;
            while (j < k) {
                int sum3 = numbers[i] + numbers[j] + numbers[k];
                if (sum3 < 0) {
                    j++;
                    while (j<k && numbers[j] == numbers[j-1]) j++;
                } else if (sum3 > 0) {
                    k--;
                    while (j<k && numbers[k] == numbers[k+1]) k--;
                } else {
                    res.push_back({numbers[i], numbers[j], numbers[k]});
                    // break; // for result can be [[-2,-1,3], [-2,1,1]]
                    j++;
                    while (j<k && numbers[j] == numbers[j-1]) j++;
                    k--;
                    while (j<k && numbers[k] == numbers[k+1]) k--;
                }
            }
            i++;
            while (i<n-2 && numbers[i] == numbers[i-1]) i++;
        }
        return res;
    }
};