Description
Given an integer array nums, return all the triplets [nums[i], nums[j], nums[k]] such that i != j, i != k, and j != k, and nums[i] + nums[j] + nums[k] == 0.
Notice that the solution set must not contain duplicate triplets.
Example 1:
Input: nums = [-1,0,1,2,-1,-4] Output: [[-1,-1,2],[-1,0,1]] Explanation: nums[0] + nums[1] + nums[2] = (-1) + 0 + 1 = 0. nums[1] + nums[2] + nums[4] = 0 + 1 + (-1) = 0. nums[0] + nums[3] + nums[4] = (-1) + 2 + (-1) = 0. The distinct triplets are [-1,0,1] and [-1,-1,2]. Notice that the order of the output and the order of the triplets does not matter.
Example 2:
Input: nums = [0,1,1] Output: [] Explanation: The only possible triplet does not sum up to 0.
Example 3:
Input: nums = [0,0,0] Output: 0,0,0 Explanation: The only possible triplet sums up to 0.
Constraints:
3 <= nums.length <= 3000-105 <= nums[i] <= 105
Code
要注意 skip duplicate 的順序：while(i + 1 < n && nums[i] == nums[i + 1]) i++;要擺在最後。
class Solution {
public:
vector<vector<int>> threeSum(vector<int>& nums) {
sort(nums.begin(), nums.end());
int n = nums.size();
vector<vector<int>> res;
for(int i = 0; i < n; i++) {
int target = -nums[i];
int l = i + 1, r = n - 1;
while(l < r) {
if(nums[l] + nums[r] == target) {
res.push_back({nums[i], nums[l], nums[r]});
int left = nums[l], right = nums[r];
// skip duplicate
while(l < r && nums[l] == left)
l++;
while(l < r && nums[r] == right)
r--;
} else if(nums[l] + nums[r] > target) {
r--;
} else if(nums[l] + nums[r] < target) {
l++;
}
}
// skip duplicate of i
// case like [-1, -1, -1, 2, ...]
// i should point to the last
// -1 after this while loop, and then the for loop
// will increment it by 1, which points to 2
while(i + 1 < n && nums[i] == nums[i + 1])
i++;
}
return res;
}
};