文章目录
一、题目
15. 3Sum
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
题目来源： leetcode
二、题解
利用双指针思路解题，关键在于去重，对i，left，right分别进行去重。
cpp
class Solution {
public:
vector<vector<int>> threeSum(vector<int>& nums) {
int n = nums.size();
sort(nums.begin(),nums.end());
vector<vector<int>> res;
for(int i = 0;i < n;i++){
if(nums[i] > 0) return res;
if(i > 0 && nums[i] == nums[i-1]) continue;
int left = i + 1,right = n - 1;
while(left < right){
int tmp = nums[i] + nums[left] + nums[right];
if(tmp > 0) right--;
else if(tmp < 0) left++;
else{
vector<int> ve;
ve.push_back(nums[i]);
ve.push_back(nums[left]);
ve.push_back(nums[right]);
res.push_back(ve);
while(left < right && nums[left] == nums[left+1]) left++;
while(left < right && nums[right-1] == nums[right]) right--;
left++;
right--;
}
}
}
return res;
}
};