3 Sum

出处

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)

Solution

Suppose the input array is S[0..n-1]. 3SUM can be solved in O(n² ) time on average by inserting each number S[i] into a hash table, and then for each index i and j, checking whether the hash table contains the integer -(S[i]+S[j]).

A better solution is using two pointers instead of one. This makes time complexity of O(n^2).

To avoid duplicate, we can take advantage of sorted arrays, i.e., move pointers by >1 to use same element only once.

Complexity

时间复杂度 O(n² )，空间复杂度 O(n)

Code

public ArrayList<ArrayList<Integer>> threeSum(int[] num) {
    ArrayList<ArrayList<Integer>> result = new ArrayList<ArrayList<Integer>>();
 
    if (num.length < 3)
        return result;
 
    // sort array
    Arrays.sort(num);
 
    for (int i = 0; i < num.length - 2; i++) {
        // //avoid duplicate solutions
        if (i == 0 || num[i] > num[i - 1]) {
 
            int negate = -num[i];
 
            int start = i + 1;
            int end = num.length - 1;
 
            while (start < end) {
                //case 1
                if (num[start] + num[end] == negate) {
                    ArrayList<Integer> temp = new ArrayList<Integer>();
                    temp.add(num[i]);
                    temp.add(num[start]);
                    temp.add(num[end]);
 
                    result.add(temp);
                    start++;
                    end--;
                    //avoid duplicate solutions
                    while (start < end && num[end] == num[end + 1])
                        end--;
 
                    while (start < end && num[start] == num[start - 1])
                        start++;
                //case 2
                } else if (num[start] + num[end] < negate) {
                    start++;
                //case 3
                } else {
                    end--;
                }
            }
 
        }
    }
 
    return result;
}