PROBLEM:

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]
]

SOLVE:

class Solution {
public:
    vector<vector<int>> threeSum(vector<int>& nums) {
        vector<vector<int>> res;
        if(nums.size()<3) // 跳過重複的
            return res;
        sort(nums.begin(),nums.end()); // 排序
        for(int i=0;i<nums.size()-2;i++){
            if(i>0&&nums[i]==nums[i-1]){
                continue;
            }
            int target=0-nums[i];
            int head=i+1;
            int tail=nums.size()-1;
            // 變為求兩個數和為target的種類
            while(head<tail){
                if(target>nums[head]+nums[tail]){
                    head++;
                }
                else if(target<nums[head]+nums[tail]){
                    tail--;
                }
                else{
                    vector<int> unit(3);
                    unit[0]=nums[i];
                    unit[1]=nums[head];
                    unit[2]=nums[tail];
                    res.push_back(unit);
                    // 跳過重複的
                    while(head<tail&&nums[head]==unit[1])head++;
                    while(head<tail&&nums[tail]==unit[2])tail--;
                }
            }
        }
        return res;
    }
};