題目敘述：https://leetcode.com/problems/3sum/
解題想法：Two Pointers 【筆記】
- 沒加上 Line-5 時，會RE (AddressSanitizer: heap-buffer-overflow)，不懂為什麼？
- 題目要求：Find all unique triplets in the array which gives the sum of zero. 因此加上 Line-12/24/25，避免數字重複的組合。
class Solution {
public:
vector<vector<int>> threeSum(vector<int>& nums) {
vector<vector<int> > ans;
if (nums.size() < 3 || (nums.size()==3 && nums[0]+nums[1]+nums[2]!=0)) return ans;
vector<int> v;
sort(nums.begin(), nums.end());
int l = 0;
int r = nums.size() - 1;
for (int i=0; i<nums.size()-2; i++) {
if (i==0 || (i>0 && nums[i]!=nums[i-1])){
int l = i + 1;
int r = nums.size() - 1;
while (l < r){
if (nums[i] + nums[l] + nums[r] > 0) r--;
else if (nums[i] + nums[l] + nums[r] < 0) l++;
else{
v.clear();
v.push_back(nums[i]);
v.push_back(nums[l]);
v.push_back(nums[r]);
ans.push_back(v);
while (l<r && nums[l]==v[1]) l++;
while (l<r && nums[r]==v[2]) r--;
}
}
}
}
return ans;
}
};