### leetcode Question 1: 3 sum

Now I started to practice the questions from a very useful website: leetcode.

There are several classic coding exercises with a well-build online-judge interface.

The code provided below can be tested using the online-judge interface here.

Q1: 3SUM.

Given an array S of n integers, are there elements abc 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.
    For example, given array S = {-1 0 1 2 -1 -4},

A solution set is:
(-1, 0, 1)
(-1, -1, 2)

Analysis:

1. The easiest way of this problem is search every 3 element pairs, after sorting the input array.
It has the complexity O(n^3).

The key point is as follows:

The problem requires output that are unique pairs. We can use a flag to test if the current element is the same as the one in previous iteration, if so, go to the next one and check again. But remember do not check the element in the first round of its own iteration.(which allows the possibility that 3 elements are equal.)
e.g. 0, 0, 0, 0,
In the first round:
1st  = [0];
2nd = 1st+1 = 0;
3rd = 2nd+1 = 0;
1st+2nd+3rd = 0, find one answer;
here 3rd goes to next element in the array, which is 0, equal to the previous one, so go to the next(which is the end of the vector), or there will be a duplicate answer.

2. The more efficient way.  O(n^2)

After sorting the input array, 3 three pointers are initialized.
The first pointer goes from index 1 to index end-2.
The 2nd pointers goes from the current element(1st pointer) index + 1
The 3rd pointers goes from the last element in a reverse order.

Two iterations:
1. 1st pointer from 1 to index end-2
2. while (2nd<3rd)

Conditions:
1. if (array[1st]+array[2nd]+array[3rd]==0), get one result
2. if (array[1st]+array[2nd]+array[3rd]>0), 3rd -1
3. if (array[1st]+array[2nd]+array[3rd]<0), 2nd +1

Code Misc:
1. vector constructor:
int myint[] = {1,2,3,4,5};
vector<int> vec(myint, myint + sizeof(myint)/sizeof(int));

2. vector.at()
This is basically the same as the [] operator, but .at() check the bounds of valid element and through a  exception if it is not.

3. vector.end().   This is the post-last-element, is not the last element, conceptually is the element that follow      the last element.

4. iterator.  * is used to get the value of the iterator. e.g. if(*it1 + *it2 + *it3  == 0)

The code of O(n^3) method:
class Solution {
public:
vector< vector<int> > threeSum(vector<int> &num) {
// Start typing your C/C++ solution below
// DO NOT write int main() function
vector<int> res_m(3);
vector< vector<int> > res;
vector&ltint&gt::iterator it1,it2,it3,iend;

sort(num.begin(),num.end());
if(num.size()>=3){
iend = num.end();
it1 = num.begin();
bool f1 = false;
for ( ;it1!=iend;++it1){
if (f1 && (*it1==*(it1-1))){ continue;}
f1 = true;
bool f2 =false;
for (it2 = it1+1;it2!=iend;++it2){
if (f2 && (*it2==*(it2-1))){ continue;}
f2 = true;
bool f3 = false;
for (it3 = it2+1;it3!=iend;++it3){
if (f3 && (*it3==*(it3-1))){continue;}
f3 = true;
if (*it1 + *it2 + *it3==0){
res_m.at(0)=*it1;
res_m.at(1)=*it2;
res_m.at(2)=*it3;
res.push_back(res_m);
break;
}

}
}
}
}
return res;
}
};


The code of O(n^2) method:
class Solution {
public:
vector<vector <int> > threeSum(vector<int> &num) {
// Start typing your C/C++ solution below
// DO NOT write int main() function
vector<int> res_m(3);
vector< vector<int> > res;
vector<int>::iterator it1,it2,it3,iend;

sort(num.begin(),num.end());
if(num.size()>=3){
iend = num.end();
it1 = num.begin();
bool f1 = false;
for ( ;it1!=iend-2;++it1){
if(f1 && *it1==*(it1-1)) {continue;}
f1 = true;
bool f2 = false;
bool f3 = false;
it3 = iend-1;
it2 = it1+1;
while (it2!=it3){

if(f3 && *it3==*(it3+1)) { it3 = it3 -1 ;continue;}
if(f2 && *it2==*(it2-1)) { it2 = it2 +1 ;continue;}

if (*it1 + *it2 + *it3==0){
res_m.at(0)=*it1;
res_m.at(1)=*it2;
res_m.at(2)=*it3;
res.push_back(res_m);
}
if (*it1 + *it2 + *it3 >0){
it3=it3-1;
f3 = true;

}else{
it2=it2+1;
f2 = true;
}

}

}
}
return res;
}
};