15.3Sum (Two-Pointers)

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:

• Elements in a triplet (a,b,c) must be in non-descending order. (ie, a ≤ b ≤ c)
• The solution set must not contain duplicate triplets.

思路：以下解法，固定两个数，用二分法查找第三个数，时间复杂度O(n2logn)，结果超时

class Solution {

public:

    vector<vector<int>> threeSum(vector<int>& nums) {

        vector<vector<int>> result;

        vector<int> item;

        int i, j, target;

        int size = nums.size();

        if(size < 3) return result;

        

        sort(nums.begin(), nums.end());

        for(i = 0; i < size-2; i++){

            for(j = i+1; j < size-1; j++){

                target = 0-nums[i]-nums[j];

                if(binarySearch(nums, j+1, size-1, target)){

                    item.clear();

                    item.push_back(nums[i]);

                    item.push_back(nums[j]);

                    item.push_back(target);

                    result.push_back(item);

                }

            }

        }

        return result;

    }

    bool binarySearch(vector<int>& nums, int start, int end, int target){

        if(start == end) return (nums[start]==target?true:false);

        int mid = (start + end) >> 1;

        if(nums[mid] == target) return true;

        else if(nums[mid] < target) return binarySearch(nums, mid+1, end, target);

        else return binarySearch(nums, start, mid, target);

    }

};

以下方法OK，时间复杂度O(n2)。需要注意跳过重复的数字

class Solution {

public:

    vector<vector<int>> threeSum(vector<int>& nums) {

        int size = nums.size();

        if(size < 3) return result;

        

        sort(nums.begin(), nums.end());

        find(nums, 1, size-1, -nums[0]);

        for(int i = 1; i < size-2; i++){

            if(nums[i]!=nums[i-1]) find(nums, i+1, size-1, -nums[i]);

        }

        return result;

    }

    void find(vector<int>& nums, int start, int end, int target){

        int sum;

        while(start<end){

            sum = nums[start]+nums[end];

            if(sum == target){

                item.clear();

                item.push_back(-target);

                item.push_back(nums[start]);

                item.push_back(nums[end]);

                result.push_back(item);

                do{ 

                    start++;

                }while(start!= end && nums[start] == nums[start-1]);

                do{ 

                    end--;

                }while(end!=start && nums[end] == nums[end+1]);

            }

            else if(sum>target){

                do{ 

                    end--;

                }while(end!=start && nums[end] == nums[end+1]);

            }

            else{

                do{ 

                    start++;

                }while(start!= end && nums[start] == nums[start-1]);

            }

        }

    }

    

private:

    vector<vector<int>> result;

    vector<int> item;

};