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：首先大循环第一个数，从第一个数之后首尾指针向中间夹逼，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;
};