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)

java:

public class Solution {    public List<List<Integer>> threeSum(int[] num) {        int len = num.length;        List<List<Integer>> list = new LinkedList<List<Integer>>();        if(len<3){            return list;        }        Arrays.sort(num);        int i=0;        while(i<len-2){            int s=i+1;            int e=len-1;            while(s<e){                if(num[s]+num[e]==0-num[i]){                    List<Integer> lst = new LinkedList<Integer>();                    lst.add(num[i]);                    lst.add(num[s]);                    lst.add(num[e]);                    list.add(lst);                    while(s+1<len&&num[s+1]==num[s]){                        s++;                    }                    s++;                    while(e-1>=0&&num[e-1]==num[e]){                        e--;                    }                    e--;                }else if(num[s]+num[e]<0-num[i]){                    s++;                }else{                    e--;                }            }            while(i+1<len&&num[i+1]==num[i]){                i++;            }            i++;        }        return list;    }}

c++:

class Solution {public:    vector<vector<int> > threeSum(vector<int> &num) {        vector<vector<int> > r;        int n=num.size();        sort(num.begin(),num.end());        if(n<3)            return r;                int i=0;        while(i<n-2){            int k1 = num[i];            int s = i+1,e=n-1;            while(s<e){                if(num[s]+num[e]==0-k1){                    vector<int> v;                    v.push_back(k1);                    v.push_back(num[s]);                    v.push_back(num[e]);                                        r.push_back(v);                    while(s+1<=n-1&&num[s]==num[s+1]){                        s++;                    }                    s++;                    while(e-1>=i&&num[e]==num[e-1]){                        e--;                    }                    e--;                }else if(num[s]+num[e]<0-k1){                    s++;                }else{                    e--;                }            }            while(i+1<=n-1&&num[i]==num[i+1]){                i++;            }            i++;        }        return r;    }};