3 Sum leetcode java

题目：

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)

题解：
 3 Sum是two Sum的变种，可以利用two sum的二分查找法来解决问题。
 本题比two sum增加的问题有：解决duplicate问题，3个数相加返回数值而非index。
 首先，对数组进行排序。
 然后，从0位置开始到倒数第三个位置（num.length-3)，进行遍历，假定num[i]就是3sum中得第一个加数，然后从i+1的位置开始，进行2sum的运算。
 当找到一个3sum==0的情况时，判断是否在结果hashset中出现过，没有则添加。(利用hashset的value唯一性）
 因为结果不唯一，此时不能停止，继续搜索，low和high指针同时挪动。

 时间复杂度是O(n2)
 实现代码为：

 1     public ArrayList<ArrayList<Integer>> threeSum(int[] num) {
 2         ArrayList<ArrayList<Integer>> res = new ArrayList<ArrayList<Integer>>();
 3         if(num.length<3||num == null)
 4             return res;
 5         
 6         HashSet<ArrayList<Integer>> hs = new HashSet<ArrayList<Integer>>();
 7         
 8         Arrays.sort(num);
 9         
         for(int i = 0; i <= num.length-3; i++){
             int low = i+1;
             int high = num.length-1;
             while(low<high){//since they cannot be the same one, low should not equal to high
                 int sum = num[i]+num[low]+num[high];
                 if(sum == 0){
                     ArrayList<Integer> unit = new ArrayList<Integer>();
                     unit.add(num[i]);
                     unit.add(num[low]);
                     unit.add(num[high]);
                     
                     if(!hs.contains(unit)){
                         hs.add(unit);
                         res.add(unit);
                     }
                     
                     low++;
                     high--;
                 }else if(sum > 0)
                     high --;
                  else
                     low ++;
             }
         }
         
         return res;
     }

 同时，解决duplicate问题，也可以通过挪动指针来解决判断，当找到一个合格结果时，将3个加数指针挪动到与当前值不同的地方，才再进行继续判断，代码如下：

 1     public ArrayList<ArrayList<Integer>> threeSum(int[] num) {
 2         ArrayList<ArrayList<Integer>> res = new ArrayList<ArrayList<Integer>>();
 3         if(num.length<3||num == null)
 4             return res;
 5         
 6         Arrays.sort(num);
 7         
 8         for(int i = 0; i <= num.length-3; i++){
 9             if(i==0||num[i]!=num[i-1]){//remove dupicate
                 int low = i+1;
                 int high = num.length-1;
                 while(low<high){
                     int sum = num[i]+num[low]+num[high];
                     if(sum == 0){
                         ArrayList<Integer> unit = new ArrayList<Integer>();
                         unit.add(num[i]);
                         unit.add(num[low]);
                         unit.add(num[high]);
                         
                         res.add(unit);
                         
                         low++;
                         high--;
                         
                         while(low<high&&num[low]==num[low-1])//remove dupicate
                             low++;
                         while(low<high&&num[high]==num[high+1])//remove dupicate
                             high--;
                             
                     }else if(sum > 0)
                         high --;
                      else
                         low ++;
                 }
             }
         }
         return res;
     }