15. 3Sum(字典) (双指针)

 

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]
]
 

class Solution {
public:
    vector<vector<int>> threeSum(vector<int>& nums) {
        
        sort(nums.begin(),nums.end());
        vector<vector<int>> res;
        if (nums.size() < 3) return res;
        for (int i = 0;i < nums.size()-1;++i) {
            if(i>0 && nums[i]==nums[i-1]) continue;
            int low = i+1;
            int high = nums.size() -1;
            while(low < high) {
                int sum = nums[i] + nums[low] + nums[high];
                if (sum == 0) {
                    res.emplace_back(vector<int>({nums[i],nums[low],nums[high]}));
                    while(low<high && nums[low+1]== nums[low]) low++;
                    while(low<high && nums[high-1]== nums[high]) high--;
                    low++;
                    high--;
                } else if (sum > 0) {
                    high--;
                } else if (sum < 0) {
                    low++;
                }
            }
        }
        return res;
    }
};

 

借助 2sum中字典的应用。  时间复杂度 o(n**2)

class Solution(object):
    def threeSum(self, nums):
        """
        :type nums: List[int]
        :rtype: List[List[int]]
        """
        if len(nums) < 3:
            return []
        nums.sort()
        res = set()
        for i, v in enumerate(nums[:-2]):
            d = {}
            for x in nums[i+1:]:
                if x not in d:
                    d[-v-x] = 1
                else:
                    res.add((v, -v-x, x))
        return map(list, res)