3Sum

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)

这题就是一个bianry search的变种。我们在固定一个元素后，相当于做一个two sum。有一个test case需要注意的是[0,0,0,0]。 所以我们需要略过这种重复多个元素的情况。

    public List<List<Integer>> threeSum(int[] num) {
        List<List<Integer>> res = new ArrayList<List<Integer>>();
        if (num == null || num.length == 0) {
            return res;
        }
        Arrays.sort(num);
        for (int i = 0; i < num.length - 2; i++) {
            while (i != 0 && i < num.length && num[i] == num[i - 1] ) {
                i++;
            }
            /*   if (i != 0 && num[i] == num[i - 1]) {
                continue; // to skip duplicate numbers; e.g [0,0,0,0]
            }

*/
            int start = i + 1;
            int end = num.length - 1;
            while (start < end) {
                int val = num[i] + num[start] + num[end];
                if (val == 0) {
                    ArrayList<Integer> subRes = new ArrayList<Integer>();
                    subRes.add(num[i]);
                    subRes.add(num[start]);
                    subRes.add(num[end]);
                    res.add(subRes);
                    do{
                        start++;
                    }while (start < end && num[start] == num[start - 1]);
                    do {
                        end--;
                    }while (start < end && num[end] == num[end + 1]);
                }else if (val < 0) {
                    start++;
                }else if (val > 0) {
                    end--;
                }
                
            }
        }
        return res;
    }