57. 三数之和 &&

题目

57. 三数之和

给出一个有n个整数的数组S，在S中找到三个整数a, b, c，找到所有使得a + b + c = 0的三元组。
样例

如S = {-1 0 1 2 -1 -4}, 你需要返回的三元组集合的是：

(-1, 0, 1)

(-1, -1, 2)
注意事项

在三元组(a, b, c)，要求a <= b <= c。

结果不能包含重复的三元组。

解析

• 主要注意去除重复元素的方法

class Solution_57 {
public:
	/**
	* @param numbers: Give an array numbers of n integer
	* @return: Find all unique triplets in the array which gives the sum of zero.
	*/

	vector<vector<int>> threeSum(vector<int> &numbers) {
		// write your code here
		vector<vector<int>> ret;
		if (numbers.size() < 3)
			return ret;
		sort(numbers.begin(), numbers.end());
		vector<int> vec;
		for (int i = 0; i < numbers.size() - 2; i++)
		{
			if (i >= 1 && numbers[i - 1] == numbers[i]) //去重复元素
			{
				continue;
			}

			int b = i + 1;
			int c = numbers.size() - 1;

			while (b<c)
			{
				while (b<c&&c<numbers.size() - 1 && numbers[c + 1] == numbers[c]) //去重只需要在查找到过后进行，所以放在下面else里面，容易理解一些
					c--;
				while (b<c&&b - 1>i&&numbers[b - 1] == numbers[b])
					b++;
				if (b == c)
					continue;

				int sum = numbers[i] + numbers[b] + numbers[c];
				if (sum>0)
				{
					c--;
				}
				else if (sum < 0)
				{
					b++;
				}
				else
				{
					vec.clear();
					vec.push_back(numbers[i]);
					vec.push_back(numbers[b]);
					vec.push_back(numbers[c]);
					ret.push_back(vec);
					//break; //bug
					c--;
					b++;
				}
			}
		}
		return ret;
	}


	vector<vector<int>> threeSum(vector<int> &numbers) {
		// write your code here
		vector<vector<int>> ret;
		if (numbers.size() < 3)
			return ret;
		sort(numbers.begin(), numbers.end());
		vector<int> vec;
		for (int i = 0; i < numbers.size() - 2; i++)
		{
			if (i >= 1 && numbers[i - 1] == numbers[i]) //去重复元素
			{
				continue;
			}

			int b = i + 1;
			int c = numbers.size() - 1;

			while (b<c)
			{

				int sum = numbers[i] + numbers[b] + numbers[c];
				if (sum>0)
				{
					c--;
				}
				else if (sum < 0)
				{
					b++;
				}
				else
				{
					vec.clear();
					vec.push_back(numbers[i]);
					vec.push_back(numbers[b]);
					vec.push_back(numbers[c]);
					ret.push_back(vec);
					//break; //bug
					c--;
					b++;

					while (b < c&&numbers[c + 1] == numbers[c])
						c--;
					while (b < c&&numbers[b - 1] == numbers[b])
						b++;

				}
			}
		}
		return ret;
	}
};