1 題目要求：

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]
]

2 解決方法:

類比於sum two，很容易想到將第三個數的相反數作為兩數之和，得到以下程式碼，但是最後一個測試的時間超時。

class TreeSumLess {
public:
	bool operator()(const vector<int>&nums1, const vector<int>&nums2) {
		int i = 0;
		while ((i < 3) && (nums1[i] == nums2[i])) i++;					
		if (i == 3) return false;
		else if (nums1[i] < nums2[i]) return true;
		else return false;
	}
};
class Solution {
public:
	vector<vector<int>> threeSum(vector<int>& nums) {
		vector<vector<int>> ret;
		for (int i = 0; i < nums.size(); i++) {
			unordered_map<int, int> hash;
			for (int j = i+1; j < nums.size(); j++) {
				int value = -nums[i] - nums[j];
				unordered_map<int, int>::iterator it1 = hash.find(value);
				if (it1 != hash.end()) {
					vector<int> temp = {nums[i],nums[j],value};				
					sort(temp.begin(), temp.end(), less<int>());
					ret.push_back(temp);
				}
				hash[nums[j]] = value;
			}
		}
		sort(ret.begin(), ret.end(), TreeSumLess());
		vector<vector<int>>::iterator pos = unique(ret.begin(), ret.end());
		/*ret.resize(distance(ret.begin(), pos));*/
		ret.erase(pos,ret.end());
		return ret;
	}
};
 

在特定情況下，例如全0，排序部分刪重過程計算量很大，進一步改進的演算法：

先對向量進行排序，中間可以根據排序特性減少重複的計算。

class Solution {
public:
	vector<vector<int> > threeSum(vector<int> &num) {
		vector<vector<int> > res;
		sort(num.begin(), num.end());
		for (int i = 0; i < num.size(); i++) {
			int target = -num[i];
			int front = i + 1;
			int back = num.size() - 1;
			while (front < back) {
				int sum = num[front] + num[back];
				// Finding answer which start from number num[i]
				if (sum < target)
					front++;
				else if (sum > target)
					back--;
				else {
					vector<int> triplet(3, 0);
					triplet[0] = num[i];
					triplet[1] = num[front];
					triplet[2] = num[back];
					res.push_back(triplet);
					// Processing duplicates of Number 2
					// Rolling the front pointer to the next different number forwards
					while (front < back && num[front] == triplet[1]) front++;
					// Processing duplicates of Number 3
					// Rolling the back pointer to the next different number backwards
					while (front < back && num[back] == triplet[2]) back--;
				}
			}
			// Processing duplicates of Number 1
			while (i + 1 < num.size() && num[i + 1] == num[i])
				i++;
		}
		return res;
	}
};
 

演算法複雜度O(n2)

空間複雜度O(1)

3 總結

複習了雜湊表、vector中sort erase、unique。排序預處理後的資料，在演算法中會增加順序的特性可以利用。: