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)
这道题目脑子没转过来，一直想要用贪心的方式做，然后各种bug，后来转变过来，可以根据target来选择。还有一种是可以用暴力方式做，这种方法一个很烦的地方就是去除重复的子项。自己试过好多中方式，然后发现自己对hashCode方法没有理解彻底，特别在重写的hashcode方法不知道。

 public int hashCode() {
        int hashCode = 1;
        for (E e : this)
            hashCode = 31*hashCode + (e==null ? 0 : e.hashCode());
        return hashCode;
    }

这个是在AbstractList这个类中重写的方法。所以我们可以用HashSet来排除多余项。具体代码都是网上copy下来的，另一种方法的详细解释在这里这里
暴力方法代码

 public List<List<Integer>> threeSum(int[] num) {
	        HashSet rs = new HashSet();
	        
	        int len = num.length;
	        Arrays.sort(num);
	        if(len <= 2) return new ArrayList(rs);
	        
	        for(int i = 0; i < len-2; i++) {
	            if(num[i] > 0) break;
	            for(int k = len-1; k > i+1; k--) {
	                if(num[k] < 0) break;
	                int ab = num[i] + num[k];
	                int c = -ab;
	                int j = bs(c, num, i+1, k-1);
	                if(j>0) {
	                    ArrayList elem = new ArrayList();
	                    elem.add(num[i]);
	                    elem.add(num[j]);
	                    elem.add(num[k]);
	                   elem.hashCode();
	                    rs.add(elem);
	                }
	            }
	        }
	        return new ArrayList(rs);
	    }
	    int bs(int c, int[] num, int l, int r) {
	        if(num[l] > c || num[r] < c) return -1;
	        while(l <= r) {
	            int m = (l+r)/2;
	            if(num[m] == c) return m;
	            else if(num[m] < c) l = m+1;
	            else r = m-1;
	        }
	        return -1;
	    }

另外的方式

	    public ArrayList<ArrayList<Integer>> threeSum(int[] num) {
        ArrayList<ArrayList<Integer>> result = new ArrayList<ArrayList<Integer>>();
        if (num == null || num.length < 3)
            return result;
        int n = num.length;
        Arrays.sort(num);
        for (int i = 0; i < n; i++) {
            int target = -num[i];
            int p = i + 1, q = n - 1;
            while (p < q) {
                if (num[p] + num[q] < target)
                    p++;
                else if (num[p] + num[q] > target)
                    q--;
                else {
                    ArrayList<Integer> tmp = new ArrayList<Integer>();
                    tmp.add(-target);
                    tmp.add(num[p]);
                    tmp.add(num[q]);
                    result.add(tmp);
                    p++;
                    q--;
                    // remove duplicates
                    while (p < n && num[p] == num[p - 1])
                        p++;
                    while (q >= i + 1 && num[q] == num[q + 1])
                        q--;
 
                }
 
            }
            // remove duplicates
            while (i < n - 1 && num[i + 1] == num[i])
                i++;
 
        }
 
        return result;
    }