LeetCode15. 3Sum

LeetCode15. 3Sum

題目：

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: 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]
]

題目分析：

先給數組排序，然後固定頭部，然後還是一個2Sum的問題，但是不一樣的是，這個數組是排序的，所以可以根據和target的大小不同來控制begin，end的不同。

if (nums[i] + nums[begin] + nums[end] < 0) {
	begin++;
}
else if (nums[i] + nums[begin] + nums[end] > 0) {
	end--;
}

else if (nums[i] + nums[begin] + nums[end] == 0) {
	tmp.push_back(nums[i]);
	tmp.push_back(nums[begin]);
	tmp.push_back(nums[end]);
	kset.insert(tmp);
	tmp.clear();
	begin++;
}

代碼：

class Solution {
public:
	vector<vector<int>> threeSum(vector<int>& nums) {
		vector<vector<int>> res;
		set<vector<int>> kset;
		vector<int> tmp;
		if (nums.empty()||nums.size()<3) return res;
		sort(nums.begin(), nums.end());
		int begin, end;
		for (int i = 0; i < nums.size()-2; i++) {
			begin = i + 1, end = nums.size() - 1;
			while (begin < end) {
				if (nums[i] + nums[begin] + nums[end] < 0) {
					begin++;
				}
				else if (nums[i] + nums[begin] + nums[end] > 0) {
					end--;
				}
				else if (nums[i] + nums[begin] + nums[end] == 0) {
					tmp.push_back(nums[i]);
					tmp.push_back(nums[begin]);
					tmp.push_back(nums[end]);
					kset.insert(tmp);
					tmp.clear();
					begin++;
				}
			}
		}
		for (auto i : kset) res.push_back(i);
		return res;
	}
};