15. 3Sum-优快云博客

该编程题目要求在给定整数数组中找到所有满足条件的不重复三元组(i,j,k)，其元素之和等于零。解决方案采用了排序和双指针法，保证了时间复杂度。

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.

class Solution {
public:
    vector<vector<int>> threeSum(vector<int>& nums) {
        vector<vector<int>> result;
        sort(nums.begin(),nums.end());
        for(int i = 0;i < nums.size();i ++){
            if(nums[i] > 0)
                return result;
            if(i> 0 && nums[i] == nums[i-1 ])
                continue;
            int left = i+1;
            int right = nums.size()-1;
            while(left < right){
                if(nums[i] + nums[left] + nums[right] > 0)
                    right --;
                else if(nums[i] + nums[left] +nums[right] < 0 )
                    left++;
                else if(nums[i] + nums[left] + nums[right] == 0){
                    result.push_back(vector<int>{nums[i],nums[left],nums[right]});
                    while (right > left && nums[right] == nums[right - 1]) right--;
                    while (right > left && nums[left] == nums[left + 1]) left++;
                    right --;
                    left ++;

                }
            }
        }
        return result;
    }
};