LeetCode 15. 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: 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]]

answer:

class Solution {public:    vector<vector<int>> threeSum(vector<int>& nums) {        sort(nums.begin(),nums.end());        int sum = 0;        vector<vector<int>> result;        for(int i = 0; i < nums.size(); i ++){            int start = i + 1;            int end = nums.size() - 1;            vector<int> tuple;            while( start < end){                sum = nums[i] + nums[start];                if(nums[end] > 0 - sum ) end --;                else if(nums[end] < 0 - sum ) start ++;                else {                    tuple.push_back(nums[i]);                    tuple.push_back(nums[start]);                    tuple.push_back(nums[end]);                    end --;                    start ++;                    while(start < end && nums[end] == nums[end + 1]) end --;                    while(start < end && nums[start] == nums[start - 1]) start ++;                    result.push_back(tuple);                    tuple.clear();                    vector<int>(tuple).swap(tuple);                }            }            while(i < nums.size() && nums[i + 1] == nums[i]) i++;        }        return result;    }};

another solution:

public List<List<Integer>> threeSum(int[] num) {    Arrays.sort(num);    List<List<Integer>> res = new LinkedList<>();     for (int i = 0; i < num.length-2; i++) {        if (i == 0 || (i > 0 && num[i] != num[i-1])) {            int lo = i+1, hi = num.length-1, sum = 0 - num[i];            while (lo < hi) {                if (num[lo] + num[hi] == sum) {                    res.add(Arrays.asList(num[i], num[lo], num[hi]));                    while (lo < hi && num[lo] == num[lo+1]) lo++;                    while (lo < hi && num[hi] == num[hi-1]) hi--;                    lo++; hi--;                } else if (num[lo] + num[hi] < sum) lo++;                else hi--;           }        }    }    return res;}