# What's wrong with my solution ? (Output limit)

• This is my solution:

``````class Solution {
public:

int search(vector<int> & num, int target, int l, int r) {
while (l <= r) {
int m = (l + r) / 2;
if (num[m] == target) return m;
else if (num[m] < target) l = m + 1;
else r = m - 1;
}
return -1;
}

vector<vector<int> > threeSum(vector<int> &num) {
vector<vector<int> > ans;
sort(num.begin(), num.end());
int lastI, lastJ, lastK;
lastI = lastJ = lastK = 123456789;  // to judge the duplicate triplets
for (int i = 0; i < num.size(); i++) {
for (int j = i + 1; j < num.size(); j++) {
int pos = search(num, 0 - num[i] - num[j], j + 1, num.size() - 1);  // search the third data
if (pos != -1) {
if (num[i] == lastI && num[j] == lastJ && num[pos] == lastK) continue;
vector<int> a;
a.push_back(num[i]);
a.push_back(num[j]);
a.push_back(num[pos]);
ans.push_back(a);
lastI = num[i];
lastJ = num[j];
lastK = num[pos];
}
}
}
return ans;
}
};``````

• I have found the problem.
Let's see the case below:
{-4,-2,-2,-2,0,1,2,2,2,3,3,4,4,6,6}
In this case, when i == 1 and j == 2, we found one solution is {-2, -2, 4};
But when i == 2 and j == 3, we found a same solution, and the solution can not be delete, because the process to found the 2 solution is not continue.
To solve this problem, we can make num[i]'s value never repeat.
Sorry for my poor English.

``````class Solution {
public:

int search(vector<int> & num, int target, int l, int r) {
while (l <= r) {
int m = (l + r) / 2;
if (num[m] == target) return m;
else if (num[m] < target) l = m + 1;
else r = m - 1;
}
return -1;
}

vector<vector<int> > threeSum(vector<int> &num) {
vector<vector<int> > ans;
sort(num.begin(), num.end());
int lastI, lastJ, lastK;
lastI = lastJ = lastK = 123456789;  // to judge the duplicate triplets
for (int i = 0; i < num.size(); i++) {
if (i >= 1 && num[i] == num[i - 1]) continue;
for (int j = i + 1; j < num.size(); j++) {
int pos = search(num, 0 - num[i] - num[j], j + 1, num.size() - 1);  // search the third data
if (pos != -1) {
if (num[i] == lastI && num[j] == lastJ && num[pos] == lastK) continue;
vector<int> a;
a.push_back(num[i]);
a.push_back(num[j]);
a.push_back(num[pos]);
ans.push_back(a);
lastI = num[i];
lastJ = num[j];
lastK = num[pos];
}
}
}
return ans;
}
};``````

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