[Algorithm Topic Breakthrough] Double pointers - sum of three numbers (7)

Table of contents

1. Question analysis

2. Algorithm principle

3. Code writing

Write at the end:

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1. Question analysis

Question link: 15. Sum of three numbers - Leetcode

 The question is to find non-repeating triples whose sum is 0.

Note that each element of the triple has a different position, so what does it mean to not repeat?

We can look at the first example,

He found three triples, but he only returned two,

That is, triples with the same elements are considered the same triplet. (If not, the empty set is returned.)

2. Algorithm principle

The first idea is of course brute force enumeration, specifically,

Sort first, then violently enumerate, and finally use set to remove duplicates.

Then we have to think about how to optimize N3’s violent enumeration.

After sorting, there is an ordered array, then we have to think about whether to use binary or double pointers for optimization: of course, double pointers are given priority.

Let’s look at the specific solution:

Fix an i position:

We only need to quickly find the position where the sum of left position + right position is 4 through double pointers.

3. Code writing

class Solution {
public:
    vector<vector<int>> threeSum(vector<int>& nums) {
        vector<vector<int>> ans;
        sort(nums.begin(), nums.end());
        for(int i = 0; i < nums.size() - 2; i++) {
            int left = i + 1, right = nums.size() - 1;
            while(left < right) {
                int sum = nums[i] + nums[left] + nums[right];
                if(sum < 0) left++;
                else if(sum > 0) right--;
                else { // sum == 0
                    ans.push_back({nums[i], nums[left], nums[right]});
                    while(left < right && nums[left] == nums[left + 1]) left++;
                    while(left < right && nums[right] == nums[right - 1]) right--;
                    left++, right--;
                }
            }
            while(i < nums.size() - 2 && nums[i] == nums[i + 1]) i++;
        }
        return ans;
    }
};

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