[array] leetCode-15. 3Sum-Medium

leetCode-15. 3Sum-Medium

descrition

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]
]

解析

方法1

3 重循环，时间复杂度-O(n^3)，空间复杂度-O(1)

for(int i=0; i<array.size(); i++){
	for(int j=i+1; j<array.size(); j++){
		for(int k=j+1; k<array.size(); k++){
			// 检查是否合法
		}
	}
}

方法2

思考方向：是否可以减少方法 1 中的循环查找次数？？

时间复杂度-O(n^2)，空间复杂度-O(1)。

对数组进行排序，需要时间 O(nlog(n))。使用两重循环，最外层循环 a=array[i]，内层循环使用双向指针进行遍历，b 从左到右，c 从右到左，思想和 two sum 一样。（参看代码）

code

#include <iostream>
#include <vector>
#include <algorithm>
#include <unordered_map>

using namespace std;

class Solution{
public:
	vector<vector<int> > threeSum(vector<int>& nums){
		// The solution set must not contain duplicate triplets.
		vector<vector<int> > ans;
		sort(nums.begin(), nums.end()); // ascending
		for(int i=0; i<nums.size(); i++){
			int target = -nums[i];
			int ileft = i+1;
			int iright = nums.size()-1;
			while(ileft < iright){
				int sum = nums[ileft]+nums[iright];
				if( sum == target){
					// answer
					vector<int> temp(3);
					temp[0] = nums[i];
					temp[1] = nums[ileft];
					temp[2] = nums[iright];
					ans.push_back(temp);
					// skip duplicate
					while(ileft<iright && nums[ileft] == temp[1])
						ileft++;
					while(ileft<iright && nums[iright] == temp[2])
						iright--;
				}else if (sum < target){
					ileft++;
				}else{
					// sum > target
					iright--;
				}
			}

			// skip duplicate
			while((i+1)<nums.size() && nums[i+1] == nums[i])
				i++;
			// i point to the same value, after the i++ in the for loop, i will point to next value
		}

		return ans;
	}
};

int main()
{
	return 0;
}