leetcode 3Sum 难度系数 3.6

最新推荐文章于 2023-04-15 01:48:20 发布

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最新推荐文章于 2023-04-15 01:48:20 发布

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探讨了如何找出数组中所有唯一且不重复的三元组，这些三元组的元素相加等于零。该解决方案使用了排序和双指针技巧来避免重复，并确保结果按非递减顺序排列。

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Question：

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)

public class Solution {
  private ArrayList<ArrayList<Integer>> results;

	public ArrayList<ArrayList<Integer>> threeSum(int[] num) {
		results = new ArrayList<>();
		if (num.length < 3) {
			return results;
		}
		Arrays.sort(num);
		int length = num.length - 1;
		for (int i = 0; i <= length - 2; i++) {
			if (i != 0 && (num[i] == num[i - 1])) {
				continue;
			}
			subThreeSum(num, i, i + 1, length);
		}
		return results;
	}

	private void subThreeSum(int[] num, int i, int first, int last) {
		while (first < last) {
			if (num[first] + num[last] + num[i] > 0) {
				last--;
			} else if (num[first] + num[last] + num[i] < 0) {
				first++;
			} else if (num[first] + num[last] + num[i] == 0) {
				ArrayList<Integer> result = new ArrayList<Integer>();
				result.add(num[i]);
				result.add(num[first]);
				result.add(num[last]);
				results.add(result);
				first++;
				last--;
				while (first < last && num[first] == num[first - 1]) {
					first++;
				}
				while (first < last && num[last] == num[last + 1]) {
					last--;
				}
			}
		}

	}
}