15. 3Sum hash解法 vs 双指针解法

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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]]

hash解法 120ms AC

class Solution {public:    vector<vector<int>> threeSum(vector<int>& nums) {        sort(nums.begin(), nums.end());        vector<vector<int>> res;        for(int i = 0; i < nums.size(); i++){            if(i > 0 && nums[i] == nums[i - 1]) continue;            vector<vector<int>> two = twoSum(nums, i + 1, 0 - nums[i]);            if(two.size() == 0) continue;            else{                for(auto it : two){                    it.push_back(nums[i]);                    res.push_back(it);                }                            }        }        return res;    }    vector<vector<int>> twoSum(vector<int>& nums, int start, int target){        int n = nums.size();        vector<vector<int>> res;        if(start >= n - 1) return res;        unordered_map<int, int> ht;        for(int i = start; i < n; i++){            vector<int> temp;            if(i < n - 1 && nums[i] == nums[i + 1] && 2 * nums[i] == target ){                temp = {nums[i], nums[i]};                res.push_back(temp);                while(i <= n - 2 && nums[i] == nums[i + 1]) i++;            }else{                if(i > start && nums[i] == nums[i - 1]) continue;                int tofind = target - nums[i];                if(ht.find(tofind) != ht.end() ){                    temp = {nums[i], nums[ht[tofind]]};                    res.push_back(temp);                }//else 注意这个else千万不能要                ht[nums[i]] = i;            }        }        return res;    }};

双指针解法：

class Solution {public:    vector<vector<int>> threeSum(vector<int>& nums) {        int n = nums.size();        sort(nums.begin(), nums.end());        vector<vector<int>> res;        for(int i = 0; i < n; i ++){            if(i > 0 && nums[i] == nums[i - 1]) continue;            int L = i + 1, R = n - 1;            vector<vector<int>> two;            vector<int> temp;            int target = 0 - nums[i];            if(target < 0)//妙哉！            {                break;            }            while(L < R){                if(nums[L] + nums[R] < target){                    L++;                                    }else if(nums[L] + nums[R] > target){                    R--;                                    }else{                    /*vector<int> triplet(3, 0);                     triplet[0] = num[i];                    triplet[1] = num[front];                    triplet[2] = num[back];                    res.push_back(triplet);*///这样写更佳，会省一些代码和for(auto it : two)的时间                    temp = {nums[L], nums[R]};                    two.push_back(temp);                    L++;                    R--;                }                while (L > i + 1 && L < n && nums[L] == nums[L - 1]) L++;    while (R < n - 1 && R > i && nums[R] == nums[R + 1]) R--;                            }            if(two.size() == 0) continue;            else{                for(auto it : two){                    it.push_back(nums[i]);                    res.push_back(it);                }            }        }        return res;    }};

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