15. 3Sum (Medium)

https://leetcode.com/problems/3sum/

Given an array nums of n integers, are there elements a, b, c in nums 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.

Example:

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

A solution set is:
[
  [-1, 0, 1],
  [-1, -1, 2]
]

Solutions

class Solution {

    // Adopt a series of optimization operations to reduce the time cost.

    public List<List<Integer>> threeSum(int[] nums) {
        int len = nums.length;

        List<List<Integer>> ans = new ArrayList<>();
        if (len < 3) {
            return ans;
        }

        Arrays.sort(nums);

        for (int i = 0; i < len - 2; i++) {
            // nums[i - 1] is already calculated, skip nums[i] if same value with previous one
            if (i != 0 && nums[i] == nums[i - 1]) {
                continue; // optimization 1
            }

            // Scan from i + 1 because i, j plays same role which you can be understood
            // that (i,j) is equivalent with (j,i)
            int j = i + 1;
            int k = len - 1;
            while (k > j) {
                if (j != i + 1 && nums[j] == nums[j - 1]) { // optimization 2, same as 1
                    j++;
                    continue;
                }

                if (k != len - 1 && nums[k] == nums[k + 1]) { // optimization 3, same as 1
                    k--;
                    continue;
                }

                int sum = nums[i] + nums[j] + nums[k];
                if (sum > 0) {
                    k--;
                }

                if (sum < 0) {
                    j++;
                }

                if (sum == 0) {
                    ans.add(Arrays.asList(nums[i], nums[j], nums[k]));
                    k--;
                    j++;
                }
            }
        }

        return ans;
    }
}

Incorrect Solutions

class Solution {

    // This solution takes up too much resources.

    // Time Limit Exceeded

    public List<List<Integer>> threeSum(int[] nums) {
        int len = nums.length;

        List<List<Integer>> ans = new ArrayList<>();
        if (len < 3) {
            return ans;
        }

        Arrays.sort(nums);

        // keep count of the elements
        Map<Integer, Integer> map = new HashMap<>();
        for (int i = 0; i < nums.length; i++) {
            if (!map.containsKey(nums[i])) {
                map.put(nums[i], 0);
            }

            map.put(nums[i], map.get(nums[i]) + 1);
        }

        // Used to keep the temporary answers which allow duplicates
        List<List<Integer>> tmpAns = new ArrayList<>();

        for (int i = 0; i < len; i++) {

            map.put(nums[i], map.get(nums[i]) - 1);

            for (int j = i + 1; j < len; j++) {
                int k = -(nums[i] + nums[j]);
                if (!map.containsKey(k)) {
                    continue;
                }

                map.put(nums[j], map.get(nums[j]) - 1);
                if (map.get(k) > 0) {
                    List<Integer> triplet = Arrays.asList(new Integer[]{nums[i], nums[j], k});
                    triplet.sort(Comparator.comparingInt(o -> o));
                    tmpAns.add(triplet);
                }
                map.put(nums[j], map.get(nums[j]) + 1);
            }

            map.put(nums[i], map.get(nums[i]) + 1);
        }

        // Remove duplicates from the answers.
        List<String> hashes = new ArrayList<>();
        for (int i = 0; i < tmpAns.size(); i++) {
            String hash = tmpAns.get(i).toString();
            if (hashes.contains(hash)) {
                continue;
            }

            hashes.add(hash);
            ans.add(tmpAns.get(i));
        }

        return ans;
    }
}

References

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