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15. 3Sum 👍

  • Time: $O(n^2)$
  • Space: $O(|\texttt{ans}|)$
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class Solution {
 public:
  vector<vector<int>> threeSum(vector<int>& nums) {
    if (nums.size() < 3)
      return {};

    vector<vector<int>> ans;

    ranges::sort(nums);

    for (int i = 0; i + 2 < nums.size(); ++i) {
      if (i > 0 && nums[i] == nums[i - 1])
        continue;
      // Choose nums[i] as the first number in the triplet, then search the
      // remaining numbers in [i + 1, n - 1].
      int l = i + 1;
      int r = nums.size() - 1;
      while (l < r) {
        const int sum = nums[i] + nums[l] + nums[r];
        if (sum == 0) {
          ans.push_back({nums[i], nums[l++], nums[r--]});
          while (l < r && nums[l] == nums[l - 1])
            ++l;
          while (l < r && nums[r] == nums[r + 1])
            --r;
        } else if (sum < 0) {
          ++l;
        } else {
          --r;
        }
      }
    }

    return ans;
  }
};

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class Solution {
  public List<List<Integer>> threeSum(int[] nums) {
    if (nums.length < 3)
      return new ArrayList<>();

    List<List<Integer>> ans = new ArrayList<>();

    Arrays.sort(nums);

    for (int i = 0; i + 2 < nums.length; ++i) {
      if (i > 0 && nums[i] == nums[i - 1])
        continue;
      // Choose nums[i] as the first number in the triplet, then search the
      // remaining numbers in [i + 1, n - 1].
      int l = i + 1;
      int r = nums.length - 1;
      while (l < r) {
        final int sum = nums[i] + nums[l] + nums[r];
        if (sum == 0) {
          ans.add(Arrays.asList(nums[i], nums[l++], nums[r--]));
          while (l < r && nums[l] == nums[l - 1])
            ++l;
          while (l < r && nums[r] == nums[r + 1])
            --r;
        } else if (sum < 0) {
          ++l;
        } else {
          --r;
        }
      }
    }

    return ans;
  }
}

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class Solution:
  def threeSum(self, nums: list[int]) -> list[list[int]]:
    if len(nums) < 3:
      return []

    ans = []

    nums.sort()

    for i in range(len(nums) - 2):
      if i > 0 and nums[i] == nums[i - 1]:
        continue
      # Choose nums[i] as the first number in the triplet, then search the
      # remaining numbers in [i + 1, n - 1].
      l = i + 1
      r = len(nums) - 1
      while l < r:
        summ = nums[i] + nums[l] + nums[r]
        if summ == 0:
          ans.append((nums[i], nums[l], nums[r]))
          l += 1
          r -= 1
          while nums[l] == nums[l - 1] and l < r:
            l += 1
          while nums[r] == nums[r + 1] and l < r:
            r -= 1
        elif summ < 0:
          l += 1
        else:
          r -= 1

    return ans