Summing the output of a set of three digital triplet

Given an array nums n comprises integers, it determines whether there are three elements a, b, c nums such that a + b + c = 0, the condition will not be repeated to identify all of the triples.

Note: The answer can not contain duplicate triples.

For example, given the array nums = [-1, 0, 1, 2, -1, -4],

Meet the requirements set of triples as:
[
[-1, 0, 1],
[-1, -1, 2]
]

Analysis:
., Two pointer index left, right left, right from the last to find a position from the current position with an array of i, if three positions and the corresponding value is equal to 0, it is added to the list in. Are looking for a first number before processing of different index numbers each time from left to right (or right to left) to find, after you find the number of qualifying points to the next number to be processed.

mport java.util.*;

//使得a+b+c=0，输出不重复数组
public class findarray {
    public static List<List<Integer>> findarray(int[] array) {  //一个数组，两个指针
        List<List<Integer>> ret = new ArrayList<>();
        if (array == null || array.length < 3) {
            return ret;
        }
        Arrays.sort(array);
        for (int i = 0; i < array.length - 2; ) {
            int left=i+1;
            int right=array.length-1;
            while (left < right) {
                if ( array[i] + array[left] + array[right] == 0) {
                    ret.add(Arrays.asList(new Integer[]{array[i], array[left], array[right]}));
                    //移动到下一个位置，找下一组解
                    left++;
                    right--;
                    while (left < right && array[left] == array[left - 1]) { //从左向右找第一个与之前处理的数不同的数的下标

                        left++;
                    }
                    while (left < right && array[right] == array[right + 1]) {
                        right--;//从右向左找第一个与之前处理的数不同的数的下标
                    }
                } else if ((array[i] + array[left] + array[right]) < 0) {

                    left++;
                    while (left < right && array[left] == array[left - 1]) { //从左向右找第一个与之前处理的数不同的数的下标

                        left++;
                    }
                } else {
                    right--;
                    while (left < right && array[right] == array[right + 1]) {
                        right--;//从右向左找第一个与之前处理的数不同的数的下标
                    }
                }

            }
            i++;//指向下一个要处理的数
            while(i<array.length-2&&array[i]==array[i-1]){
                i++;
            }
        }
        return ret;
    }



    public static void main(String[] args) {
        findarray f=new findarray();
        int[] array={1,4,5,6,7,6,5,3,3-1,-2,-4,-6,6,0};
        System.out.println(f.findarray(array));
    }
}

operation result