LeetCode | 15. 3Sum

最新推荐文章于 2021-12-25 14:53:18 发布

1LOVESJohnny

最新推荐文章于 2021-12-25 14:53:18 发布

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本文深入探讨了经典的三数之和算法，旨在寻找数组中三个元素相加等于零的所有唯一组合。通过详细解释二分法的应用，文章提供了一种高效解决此问题的方法，并附带了完整的C++代码实现。

题目：

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]
]

思路：

经典的二分之路的开启题目。强烈推荐大家按照题目的推荐系列做题，会把这一思路的题目思考的很透彻。

代码：

class Solution {
public:
    vector<vector<int>> threeSum(vector<int>& nums) {
        vector<vector<int>> tri;
		if (nums.size() < 1)
			return tri;
		sort(nums.begin(), nums.end());
		for (int i = 0; i < nums.size(); i++)
		{
			if (i == 0 || (i > 0 && nums[i] != nums[i - 1]))
			{
				int sum = 0 - nums[i];
				int low = i + 1;
				int high = nums.size() - 1;
				while (low < high)
				{
					int rest = nums[low] + nums[high];
					if (rest == sum)
					{						
						tri.push_back(vector<int>{nums[i], nums[low], nums[high]});
						while (low < nums.size() - 1 && nums[low] == nums[low + 1])
							low++;
						while (high > 0 && nums[high] == nums[high - 1])
							high--;
						low++;
						high--;
					}
					else if (rest < sum)
						low++;
					else high--;
				}
			}
		}
		return tri;
    }
};