Identify all satisfy the condition a + b + c = 0 will not be repeated triplet.

You contains a n array of integers  nums, determines  nums whether there are three elements a, b, c, so that  A + B + C = 0? Please do not satisfy the conditions you find all duplicate triples.

Note: The answer can not contain duplicate triples.

 

Example:

Given array nums = [-1, 0, 1 , 2, -1, -4], 

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

1, to address violence

public static List<List<Integer>> threeSum(int[] nums) {
    List<List<Integer>> list = new ArrayList<>();
    if (nums == null || nums.length < 3) {
        return list;
    }
    Arrays.sort(nums);
    for (int i = 0; i < nums.length; i++) {
        for (int j = i+1; j < nums.length; j++) {
            for (int k = j+1; k < nums.length; k++) {
                if (nums[i] + nums[j] + nums[k] == 0) {
                   list.add(Arrays.asList(nums[i], nums[j], nums[k]));
                }
            }
        }
    }
    return list;
}

But so would exceed the time limit, and has not performed to re- 
2, an improved version of array traversal

List static public <List <Integer >> threeSum (int [] the nums) { 
    List <List <Integer List the ArrayList = new new >> <> (); 
    IF (nums.length the nums == null || <. 3) return List; 
    // number of all first array in ascending order 
    Arrays.sort (the nums); 
    for (int I = 0; I <nums.length; I ++) { 
        // if the current first number to be calculated is greater than 0, after certain numbers which is greater than 0 (as it has been sorted), the sum of three numbers must be greater than 0, the end of the cycle 
        IF (the nums [I]> 0) {BREAK;} 
        // first number of the three numbers of deduplication 
        IF (I> 0 && the nums [I] == the nums [-I. 1]) {Continue;} 
        // second number index address 
        int L = I +. 1; 
        // address of the third number subscript 
        R = nums.length. 1-int; 
        // second number when the number of addresses in the third address R L when the front traversing all cases 
        while (L <R){
            int sum = nums[i] + nums[L] + nums[R];
            IF (SUM == 0) { 
                List.add (Arrays.asList (the nums [I], the nums [L], the nums [R & lt])); 
                // the three digits to a second number of weight 
                while (L <R & lt && the nums [L] == the nums [L +. 1]) {L ++;} 
                // third number three to the number of weight 
                while (L <R && nums [ R] == nums [R- . 1]) {R--;} 
                // case where the sum == 0, the first number remains the same, the second and the third number can not be repeated, so that the left traverse knitting continues rearwardly L , R continue to traverse the right forward 
                L ++; 
                R--; 
            } 
            // if <specifies the number num [L] to the left small sum 0, so the address + 1 to continue to traverse the left rearward 
            else if (sum <0 ) {L ++;} 
            // if sum> 0 described right number num [R] is too great, so to continue to traverse the right address -1 forward 
            the else iF (sum> 0) {R--;} 
        } 
    } 
    return List; 
}