015.3Sum

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Solution 1: skipped

Brute force with 3 loops, time complexity O(n^3)

Solution 2: accepted 58%

Two-pointer after sorting. 

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public class Solution {
    public List<List<Integer>> threeSum(int[] nums) {
        
        int len = nums.length;
        List<List<Integer>> result = new ArrayList<List<Integer>>();
        if (nums == null || len < 3)
            return result;
        Arrays.sort(nums);
        
        for (int i = 0; i < len - 2; i++) {
            if (i == 0 || nums[i] > nums[i - 1]) {
                int j = i + 1;
                int k = len - 1;
                while (k > j) {
                    int sum = nums[i] + nums[j] + nums[k];
                    if (sum == 0) {
                        List<Integer> al = new ArrayList<>();
                        al.add(nums[i]);
                        al.add(nums[j]);
                        al.add(nums[k]);
                        result.add(al);
                        
                        j++;
                        k--;
                        while (k > j && nums[j] == nums[j - 1])
                            j++;
                        while (k > j && nums[k] == nums[k + 1])
                            k--;
                    }
                    else if (sum < 0)
                        j++;
                    else
                        k--;
                }
            }
        }
        return result;
    }
}