leetcode_15. 3Sum-求数组中三个数和为0

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]
]
把所有满足条件的三元组都要列出来！

思路：先排序，比如先把第一个元素定为nums[i]时，让后面两个元素和为(-1)*nums[i]，转化为两元素和的问题。
遇到相同的元素就跳过去！

    public List<List<Integer>> threeSum(int[] nums) {

        List<List<Integer>> my = new ArrayList<>();
        Arrays.sort(nums);
        for(int i=0;i<=nums.length-3;i++)
            if(nums[i]<=0){
                if(i>=1 && nums[i]==nums[i-1])
                    continue;               
                int j=i+1;
                int k=nums.length -1;
                while(j<k){
                    if(nums[i]+nums[j]+nums[k]==0)
                    {
                        List<Integer> one = new ArrayList<>();
                        one.add(nums[i]);
                        one.add(nums[j]);
                        one.add(nums[k]);
                        my.add(one);    
                        j++;
                        k--;
                        while(j<k && nums[j]==nums[j-1])
                            j++;
                        while(j<k && nums[k]==nums[k+1])
                            k--;
                    }

                    else {
                        boolean flag =true;
                        // 求有序数组中两个数的和为(-1)*nums[i]
                        while(flag){
                            while(j<k && nums[j]+nums[k]>(-1)*nums[i])
                                k--;
                            while(j<k && nums[j]+nums[k]<(-1)*nums[i])
                                j++;
                            if( j>=k || nums[j]+nums[k]==(-1)*nums[i])
                                flag = false;
                        }
                    }

                }   // while(j<k)


            }

        return my;
    }