15. 3Sum

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：先排序（从小到大），然后遍历数组，固定一个元素，用两个指针first和last，分别指向数组子区间第一个元素和最后一个元素。求和sum，如果sum小于0，将first++，如果sum大于0，将last--。注意指针移动过程中过滤掉重复的元素。

class Solution {
public:
    vector<vector<int>> threeSum(vector<int>& nums) {
        vector<vector<int>> result;
	sort(nums.begin(),nums.end());
	for (int i = 0; i < nums.size();i++)
	{
		if (nums[i] > 0)
			break;		
		if (i > 0 && nums[i] == nums[i - 1])
			continue;
		int first = i+1;
		int last = nums.size() - 1;
		
		while (first < last)
		{
			int sum =nums[i] +nums[first] + nums[last];
			if (sum  == 0)
			{
				vector<int> res;
				res.push_back(nums[i]);
				res.push_back(nums[first]);
				res.push_back(nums[last]);
				result.push_back(res);				
				first++;									
			    last--;
				while (nums[first] == nums[first - 1])
					first++;
				while (nums[last] == nums[last + 1])
					last--;
			}
			else if (sum < 0)
			{
				first++;
			}				
			else
			{
				last--;
			}								
		}
	}
	return result;
    }
};