Topic: Three of the number of and
Original title links: (https://leetcode-cn.com/problems/3sum/)
Of course, the method of violence is not to say, it was originally tried a double-pointer, but forgot to determine duplicate programs; plus a repeat determination after the code program, a variety of array bounds error, find the idea of big brother on the "explanations" and found that idea is about the same, but the author is a variety of error:
(first posted this question a tragic history)
In the end, just shining the code "problem solution" in the bigwigs given "step by step" into a "people's code":
class Solution {
public:
vector<vector<int>> threeSum(vector<int>& nums) {
sort(nums.begin(), nums.end());
vector<vector<int>> ans;
int n = nums.size();
if(n < 3) return ans;
for(int i = 0; i < n - 2; i++)
{
if(i == 0 || (i > 0 && nums[i] != nums[i - 1]))
{
int L = i + 1, R = n - 1;
while(L < R)
{
if(nums[i] + nums[L] + nums[R] == 0)
{
if(L == i + 1 || nums[L] != nums[L - 1])
{
vector<int> v;
v.push_back(nums[i]);
v.push_back(nums[L]);
v.push_back(nums[R]);
ans.push_back(v);
}
L++;
R--;
}
else if(nums[i] + nums[L] + nums[R] < 0)
{
L++;
}
else
{
R--;
}
}
}
}
return ans;
}
};
Digression:
Suffering two types, one that makes you strong; the other is worthless, just inviting torment
- "House of Cards"
