【LeetCode】15. 3Sum

参考TwoSum问题

15. 3Sum

Medium

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]
]

1. 首先对vector排序。

2. 依次不重复取出每一个数nums[i]（如果取出重复的数一定会导致重复的三元组）。

3. 对剩余的数，转换成TwoSum问题，target=-nums[i]。

4. 设置双指针left=i+1和right=nums.size()-1。

5. 如果nums[left]+nums[right]==target，则记录，并且跳过所有邻近的相同元素。

6. 如果大于，right--；如果小于，left++。

//https://leetcode.com/problems/3sum/discuss/7628/Share-my-C%2B%2B-solutionvery-easy-to-understand
#include<iostream>
#include<vector>
#include<map>
#include<algorithm>
using namespace std;
vector<vector<int>> threeSum(vector<int>& nums) {
	vector<vector<int>> ret;
	int n = nums.size();
	if (n < 3)
		return ret;

	sort(nums.begin(), nums.end());

	int i = 0, left = 0, right = 0, target = 0;

	for (i = 0; i < n - 2; ++i)
	{
		//nums has been sorted in ascending order,if nums[i]>0, there are must not such triplets in the rear
		//thanks for lxyscls's advice
		if (nums[i] > 0)
			break;

		if (i != 0 && nums[i] == nums[i - 1])
			continue;

		target = 0 - nums[i];
		left = i + 1;
		right = n - 1;

		while (left < right)
		{
			if (nums[right] < 0)
				break;

			if ((nums[left] + nums[right]) == target)
			{
				vector<int> temp;
				temp.push_back(nums[i]);
				temp.push_back(nums[left]);
				temp.push_back(nums[right]);
				ret.push_back(temp);

				while (left < right && nums[left + 1] == nums[left])
					++left;
				while (left < right && nums[right - 1] == nums[right])
					--right;

				++left;
				--right;
			}
			else if ((nums[left] + nums[right]) < target)
				++left;
			else
				--right;
		}
	}

	return ret;
}
int main(void) {
	int n, elem; vector<int> input;
	while (cin >> n) {
		if (!input.empty()) input.pop_back();
		for (int i = 0; i<n; i++) {
			cin >> elem;
			input.push_back(elem);
		}
		vector<vector<int> > ans = threeSum(input);
		for (int i = 0; i<int(ans.size()); i++) {
			cout << ans[i][0] << " " << ans[i][1] << " " << ans[i][2] << endl;
		}
	}
	return 0;
}