LeetCode 15： 3 Sum

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)

与Two Sum和 4 Sum类似，算法如下：

1、先对数组nums进行排序

2、对数组中的每个元素下表i，begin=i+1， end=length-1，sum = nums[i] + nums[begin] + nums[end]

sum == 0, 将三个元素组成vector，插入到结果vector中， begin++

sum < 0. begin++

sum >0  end--

要注意对重复元素的处理，避免产生重复元素。

nums[i] == nums[i-1]，

nums[begin] == nums[begin-1]

nums[end] == nums[end+1]

代码如下：

	vector<vector<int>> threeSum(vector<int>& nums) {
		vector<vector<int>> result;
		int length = nums.size();
		if (length < 3 )
			return result;
		//先排序
		sort(nums.begin(), nums.end());
		for (int i=0; i<length-2; i++)
		{
			int begin = i+1;
			int end = length -1;
			//nums[i]和nums[i-1]相等，避免重复
			if (i>0 && nums[i]==nums[i-1])
				continue;

			while(begin <end)
			{
				//nums[begin]和nums[begin-1]相等，避免重复
				if (begin>i+1 && nums[begin]== nums[begin-1])
				{
					begin++;
					continue;
				}
				//nums[end]和nums[end+1]相等，避免重复
				if (end<length-1 && nums[end] == nums[end+1])
				{
					end--;
					continue;
				}
				int sum = nums[i] + nums[begin] + nums[end];
				if (sum < 0)
				{
					begin ++;
				}else if (sum == 0)
				{
					vector<int> tmp;
					tmp.push_back(nums[i]);
					tmp.push_back(nums[begin]);
					tmp.push_back(nums[end]);
					result.push_back(tmp);
					begin++;
				}else{
					end--;
				}

			}
		}
		return result;
	}